Application of differential equations Questions in English

(A) Let the curve be $y = f(x)$. The slope of the line joining the origin $(0, 0)$ and the point $P(x, y)$ is $m_1 = \frac{y}{x}$.
The slope of the tangent at $P(x, y)$ is $m_2 = \frac{dy}{dx}$.
Since the tangent is perpendicular to the line joining the origin to $P$,we have $m_1 \times m_2 = -1$.
Therefore,$\frac{y}{x} \times \frac{dy}{dx} = -1$.
This implies $y \, dy = -x \, dx$.
Integrating both sides,we get $\int y \, dy = -\int x \, dx$,which gives $\frac{y^2}{2} = -\frac{x^2}{2} + C$.
This simplifies to $x^2 + y^2 = 2C$,which represents a circle centered at the origin.

(B) The general equation for $Simple \ Harmonic \ Motion$ is given by $x = A \cos(nt + \phi)$,where $A$ is the amplitude and $\phi$ is the phase constant.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = -An \sin(nt + \phi)$
Next,differentiate again with respect to $t$ to find the second derivative:
$\frac{d^2x}{dt^2} = -An^2 \cos(nt + \phi)$
Since $x = A \cos(nt + \phi)$,we can substitute this into the equation:
$\frac{d^2x}{dt^2} = -n^2 x$
Rearranging the terms gives the differential equation:
$\frac{d^2x}{dt^2} + n^2x = 0$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{y + 1}{x - 1}$.
Separating the variables,we get $\frac{dy}{y + 1} = \frac{dx}{x - 1}$.
Integrating both sides,$\int \frac{dy}{y + 1} = \int \frac{dx}{x - 1}$.
This gives $\ln|y + 1| = \ln|x - 1| + C$,which simplifies to $y + 1 = k(x - 1)$ where $k$ is a constant.
We are given the initial condition $y(1) = 2$.
Substituting $x = 1$ into the equation,we get $y + 1 = k(1 - 1) = 0$,which implies $y = -1$.
However,the condition specifies $y(1) = 2$,which contradicts $y = -1$.
Since the differential equation is undefined at $x = 1$ and the initial condition leads to a contradiction,there is no solution.

(B) Given differential equation is $y\frac{dy}{dx} + x = a$.
Separating the variables,we get $y \, dy = (a - x) \, dx$.
Integrating both sides,we get $\int y \, dy = \int (a - x) \, dx$.
$\frac{y^2}{2} = ax - \frac{x^2}{2} + C$,where $C$ is the constant of integration.
Multiplying by $2$,we get $y^2 = 2ax - x^2 + 2C$.
Rearranging the terms,we get $x^2 - 2ax + y^2 = 2C$.
Adding $a^2$ to both sides,we get $(x^2 - 2ax + a^2) + y^2 = a^2 + 2C$.
$(x - a)^2 + y^2 = a^2 + 2C$.
Let $k = a^2 + 2C$,then $(x - a)^2 + y^2 = k$.
This is the equation of a set of circles with center at $(a, 0)$,which lies on the $x$-axis.

(B) Given the gradient of the curve is $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.
Integrating both sides with respect to $x$,we get:
$y = \int \left( 1 - \frac{1}{x^2} \right) dx = x + \frac{1}{x} + c$.
Since the curve passes through the point $\left( 2, \frac{7}{2} \right)$,we substitute these values into the equation:
$\frac{7}{2} = 2 + \frac{1}{2} + c$.
$\frac{7}{2} = \frac{5}{2} + c \implies c = 1$.
Thus,the equation of the curve is $y = x + \frac{1}{x} + 1$.
Multiplying by $x$,we get $xy = x^2 + 1 + x$,which is $xy = x^2 + x + 1$.

(B) Given the differential equation $x \frac{d^2y}{dx^2} = 1$.
Dividing by $x$,we get $\frac{d^2y}{dx^2} = \frac{1}{x}$.
Integrating with respect to $x$,we get $\frac{dy}{dx} = \int \frac{1}{x} dx = \log x + c_1$.
Given that $\frac{dy}{dx} = 0$ when $x = 1$,we have $0 = \log(1) + c_1$,which implies $c_1 = 0$.
So,$\frac{dy}{dx} = \log x$.
Integrating again with respect to $x$,we get $y = \int \log x dx = x \log x - x + c_2$.
Given that $y = 1$ when $x = 1$,we have $1 = 1 \log(1) - 1 + c_2$,which implies $1 = 0 - 1 + c_2$,so $c_2 = 2$.
Substituting the values of $c_1$ and $c_2$,the required solution is $y = x \log x - x + 2$.

(A) The length of the normal to a curve is given by $L = |y| \sqrt{1 + (\frac{dy}{dx})^2}$.
Given that the length of the normal is equal to the radius vector $r = \sqrt{x^2 + y^2}$,we have:
$|y| \sqrt{1 + (\frac{dy}{dx})^2} = \sqrt{x^2 + y^2}$
Squaring both sides:
$y^2 (1 + (\frac{dy}{dx})^2) = x^2 + y^2$
$y^2 + y^2 (\frac{dy}{dx})^2 = x^2 + y^2$
$y^2 (\frac{dy}{dx})^2 = x^2$
Taking the square root on both sides:
$y \frac{dy}{dx} = \pm x$
Separating the variables:
$y \, dy = \pm x \, dx$
Integrating both sides:
$\int y \, dy = \pm \int x \, dx$
$\frac{y^2}{2} = \pm \frac{x^2}{2} + C$
$y^2 = \pm x^2 + 2C$
Let $2C = k$,then the equation is ${y^2} \mp {x^2} = k$,which is equivalent to ${y^2} \pm {x^2} = k$.

(C) Let $P_0$ be the initial population and $P$ be the population at time $t$.
According to the problem,$\frac{dP}{dt} = kP$,where $k$ is a constant.
Separating the variables,we get $\frac{dP}{P} = k dt$.
Integrating both sides,we have $\ln P = kt + C$.
At $t = 0$,$P = P_0$,so $\ln P_0 = C$.
Thus,$\ln P = kt + \ln P_0$,which simplifies to $\ln \left( \frac{P}{P_0} \right) = kt$.
Given that the population doubles in $5$ hours,at $t = 5$,$P = 2P_0$.
Substituting these values,$\ln \left( \frac{2P_0}{P_0} \right) = 5k$,so $k = \frac{\ln 2}{5}$.
Now,we need to find the population at $t = 25$ hours.
$\ln \left( \frac{P}{P_0} \right) = \left( \frac{\ln 2}{5} \right) \times 25 = 5 \ln 2 = \ln(2^5) = \ln 32$.
Therefore,$\frac{P}{P_0} = 32$,which means $P = 32 P_0$.
The number of bacteria will be $32$ times the original.

(D) Given the differential equation: $\frac{d^2y}{dx^2} = \frac{\ln x}{x^2}$.
Integrating both sides with respect to $x$:
$\frac{dy}{dx} = \int \frac{\ln x}{x^2} dx$.
Using integration by parts,let $u = \ln x$ and $dv = x^{-2} dx$. Then $du = \frac{1}{x} dx$ and $v = -\frac{1}{x}$.
$\frac{dy}{dx} = -\frac{\ln x}{x} - \int (-\frac{1}{x}) \cdot \frac{1}{x} dx = -\frac{\ln x}{x} + \int x^{-2} dx = -\frac{\ln x}{x} - \frac{1}{x} + C_1 = -\frac{\ln x + 1}{x} + C_1$.
Given $\frac{dy}{dx} = -1$ at $x = 1$:
$-1 = -\frac{\ln(1) + 1}{1} + C_1 \Rightarrow -1 = -1 + C_1 \Rightarrow C_1 = 0$.
So,$\frac{dy}{dx} = -\frac{\ln x + 1}{x}$.
Integrating again:
$y = -\int \frac{\ln x + 1}{x} dx$.
Let $t = \ln x$,then $dt = \frac{1}{x} dx$:
$y = -\int (t + 1) dt = -(\frac{t^2}{2} + t) + C_2 = -\frac{(\ln x)^2}{2} - \ln x + C_2$.
Given $y = 0$ at $x = 1$:
$0 = -\frac{(\ln 1)^2}{2} - \ln(1) + C_2 \Rightarrow 0 = 0 - 0 + C_2 \Rightarrow C_2 = 0$.
Thus,the solution is $y = -\frac{1}{2}(\ln x)^2 - \ln x$.

(A) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings,i.e.,$-\frac{dT}{dt} = k(T - T_{surrounding})$.
As the temperature $T$ decreases,the rate of cooling $-\frac{dT}{dt}$ also decreases.
This means the slope of the temperature-time graph should be steeper initially and become flatter as the temperature approaches the room temperature.
Curve $A$ shows this behavior,where the rate of change is initially high and decreases over time.
Therefore,graph $A$ is the correct representation.

(C) Given the slope of the curve is $\frac{dy}{dx} = x^2 - 2x$.
Integrating both sides with respect to $x$:
$y = \int (x^2 - 2x) dx = \frac{x^3}{3} - x^2 + C$.
Since the curve passes through the point $(2, 0)$,we substitute $x = 2$ and $y = 0$ into the equation:
$0 = \frac{2^3}{3} - 2^2 + C$
$0 = \frac{8}{3} - 4 + C$
$0 = \frac{8 - 12}{3} + C$
$0 = -\frac{4}{3} + C \Rightarrow C = \frac{4}{3}$.
Thus,the equation of the curve is $y = \frac{x^3}{3} - x^2 + \frac{4}{3}$.
To find the value of $y$ at $x = 0$:
$y(0) = \frac{0^3}{3} - 0^2 + \frac{4}{3} = \frac{4}{3}$.
Therefore,the point is $(0, 4/3)$.

(C) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 3x^2 + 2x + 5$.
To find the equation of the curve,we integrate both sides with respect to $x$:
$y = \int (3x^2 + 2x + 5) dx$
$y = x^3 + x^2 + 5x + C$
Since the curve passes through the point $(0, 1)$,we substitute $x = 0$ and $y = 1$ into the equation:
$1 = (0)^3 + (0)^2 + 5(0) + C$
$C = 1$
Substituting the value of $C$ back into the equation,we get:
$y = x^3 + x^2 + 5x + 1$.

(A) Given the differential equation: $\frac{dV}{dt} = -k(T - t)$.
Integrating both sides with respect to $t$:
$V(t) = \int -k(T - t) dt = k \int (T - t) d(T - t) = \frac{k(T - t)^2}{2} + C$.
At $t = 0$,the value of the equipment is the purchase value $I$,so $V(0) = I$:
$I = \frac{k(T - 0)^2}{2} + C \implies I = \frac{kT^2}{2} + C \implies C = I - \frac{kT^2}{2}$.
Substituting $C$ back into the equation for $V(t)$:
$V(t) = \frac{k(T - t)^2}{2} + I - \frac{kT^2}{2}$.
The scrap value $V(T)$ is the value at $t = T$:
$V(T) = \frac{k(T - T)^2}{2} + I - \frac{kT^2}{2} = 0 + I - \frac{kT^2}{2} = I - \frac{kT^2}{2}$.

(A) Given the differential equation: $\frac{dp(t)}{dt} = 0.5p(t) - 450 = \frac{p(t) - 900}{2}$.
Separating the variables,we get: $\int \frac{dp(t)}{p(t) - 900} = \int \frac{1}{2} dt$.
Integrating both sides: $\ln |p(t) - 900| = \frac{1}{2} t + C$.
Using the initial condition $p(0) = 850$: $\ln |850 - 900| = \frac{1}{2}(0) + C \implies C = \ln 50$.
Thus,the equation is: $\ln |p(t) - 900| = \frac{1}{2} t + \ln 50$.
We want to find $t$ when $p(t) = 0$: $\ln |0 - 900| = \frac{1}{2} t + \ln 50$.
$\ln 900 - \ln 50 = \frac{1}{2} t$.
$\ln \left( \frac{900}{50} \right) = \frac{1}{2} t$.
$\ln 18 = \frac{1}{2} t$.
$t = 2 \ln 18$.

(C) Given the differential equation: $\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200 = \frac{p(t) - 400}{2}$.
Separating the variables,we get: $\int \frac{dp(t)}{p(t) - 400} = \int \frac{1}{2} dt$.
Integrating both sides: $\ln |p(t) - 400| = \frac{t}{2} + C$.
This implies $|p(t) - 400| = e^{C} \cdot e^{t/2}$,or $p(t) - 400 = Ke^{t/2}$ where $K = \pm e^C$.
Using the initial condition $p(0) = 100$: $100 - 400 = Ke^0 \implies K = -300$.
Substituting $K$ back into the equation: $p(t) - 400 = -300e^{t/2}$.
Therefore,$p(t) = 400 - 300e^{t/2}$.

(A) Let $V$ be the volume and $r$ be the radius of the spherical rain drop. The volume is given by $V = \frac{4}{3}\pi r^3$.
The rate of change of volume is $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
According to the problem,the rate of evaporation is proportional to the surface area $S = 4\pi r^2$. Since it is evaporating,the rate of change of volume is negative: $\frac{dV}{dt} = -K(4\pi r^2)$,where $K > 0$.
Equating the two expressions for $\frac{dV}{dt}$:
$4\pi r^2 \frac{dr}{dt} = -K(4\pi r^2)$.
Dividing both sides by $4\pi r^2$ (assuming $r \neq 0$):
$\frac{dr}{dt} = -K$,which can be written as $\frac{dr}{dt} + K = 0$.

(C) The given differential equation is $D^3y - 3D^2y - 4Dy + 12y = 0$.
Substituting $y = e^{mx}$,we get $Dy = me^{mx}$,$D^2y = m^2e^{mx}$,and $D^3y = m^3e^{mx}$.
Substituting these into the equation: $m^3e^{mx} - 3m^2e^{mx} - 4me^{mx} + 12e^{mx} = 0$.
Since $e^{mx} \neq 0$,we have the characteristic equation: $m^3 - 3m^2 - 4m + 12 = 0$.
Factoring the equation: $m^2(m - 3) - 4(m - 3) = 0 \Rightarrow (m^2 - 4)(m - 3) = 0$.
This gives $(m - 2)(m + 2)(m - 3) = 0$.
The roots are $m = 2, m = -2, m = 3$.
Since $m \in N$ (natural numbers),we only consider $m = 2$ and $m = 3$.
Thus,there are $2$ such values.

(B) Given the curve $x^{2/3} + y^{2/3} = a^{2/3}$.
Differentiating with respect to $x$,we get:
$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$
$\Rightarrow x^{-1/3} + y^{-1/3} \frac{dy}{dx} = 0$
$\Rightarrow \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3}$.
For orthogonal trajectories,replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$:
$-\frac{dx}{dy} = -\left(\frac{y}{x}\right)^{1/3}$
$\Rightarrow \frac{dx}{dy} = \frac{y^{1/3}}{x^{1/3}}$
$\Rightarrow x^{1/3} dx = y^{1/3} dy$.
Integrating both sides:
$\int x^{1/3} dx = \int y^{1/3} dy$
$\Rightarrow \frac{x^{4/3}}{4/3} = \frac{y^{4/3}}{4/3} + C$
$\Rightarrow x^{4/3} - y^{4/3} = C'$ (where $C' = \frac{4}{3}C$ is a constant).
Thus,the correct option is $B$.

(C) Let the curve be $y = f(x)$. The area of the region bounded by the coordinate axes,the curve,and the ordinate at point $(x, y)$ is given by $\int_0^x y \, dx = y^3$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$y = \frac{d}{dx}(y^3)$
$y = 3y^2 \frac{dy}{dx}$
This gives two cases:
$1) y = 0$ (which is a trivial solution,the x-axis).
$2) 1 = 3y \frac{dy}{dx} \Rightarrow 3y \, dy = dx$.
Integrating both sides:
$\int 3y \, dy = \int dx$
$\frac{3y^2}{2} = x + C$
Rearranging the equation:
$y^2 = \frac{2}{3}(x + C)$
This is the standard form of a parabola $y^2 = 4a(x - h)$.

(B) Let the equation of the straight line be $y = kx + c$,where $k$ and $c$ are constants.
Then,$\frac{dy}{dx} = k$.
Substituting this into the given differential equation $\frac{dy}{dx} + x \left( \frac{dy}{dx} \right)^2 - y = 0$,we get:
$k + xk^2 - (kx + c) = 0$
$k + xk^2 - kx - c = 0$
$(k^2 - k)x + (k - c) = 0$
For this to hold for all $x$,the coefficients must be zero:
$k^2 - k = 0 \Rightarrow k(k - 1) = 0 \Rightarrow k = 0$ or $k = 1$.
If $k = 0$,then $k - c = 0 \Rightarrow c = 0$. The line is $y = 0$.
If $k = 1$,then $k - c = 0 \Rightarrow c = 1$. The line is $y = x + 1$.
Thus,there are $2$ such straight lines.

(D) Given $y = e^{4x} + 2e^{-x}$.
First derivative: $\frac{dy}{dx} = 4e^{4x} - 2e^{-x}$.
Second derivative: $\frac{d^2y}{dx^2} = 16e^{4x} + 2e^{-x}$.
Third derivative: $\frac{d^3y}{dx^3} = 64e^{4x} - 2e^{-x}$.
Now,substitute these into the expression $\frac{d^3y}{dx^3} - 13\frac{dy}{dx}$:
$= (64e^{4x} - 2e^{-x}) - 13(4e^{4x} - 2e^{-x})$
$= 64e^{4x} - 2e^{-x} - 52e^{4x} + 26e^{-x}$
$= 12e^{4x} + 24e^{-x}$
$= 12(e^{4x} + 2e^{-x})$
$= 12y$.
Therefore,$\frac{\frac{d^3y}{dx^3} - 13\frac{dy}{dx}}{y} = \frac{12y}{y} = 12$.
Thus,$K = 12$.

(A) Let $P(x, y)$ be a point on the curve. Let $PM$ be the ordinate,where $M$ is on the $x$-axis. Let the normal at $P$ meet the $x$-axis at $T$. Let $\theta$ be the angle the tangent makes with the $x$-axis. Then the normal makes an angle $\pi - \theta$ with the $x$-axis. The projection of the ordinate $PM$ on the normal $PT$ is $PN$. In the right-angled triangle $\triangle PMN$,the angle $\angle PMN = \theta$. Thus,$PN = PM \cos \theta = y \cos \theta$. Given $PN = a$,we have $y \cos \theta = a$. Since $\cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}}$ and $\tan \theta = \frac{dy}{dx} = y_1$,we get $y \frac{1}{\sqrt{1 + y_1^2}} = a$. Squaring both sides,$y^2 = a^2(1 + y_1^2)$,which gives $y_1^2 = \frac{y^2 - a^2}{a^2}$. Thus,$\frac{dy}{dx} = \frac{\sqrt{y^2 - a^2}}{a}$. Separating variables,$\int \frac{a \, dy}{\sqrt{y^2 - a^2}} = \int dx$. Integrating,we get $a \ln |y + \sqrt{y^2 - a^2}| = x + c$.

(A) Given the equation of the parabola is $y = ax^2$ ... $(1)$
Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = 2ax$.
From $(1)$,we have $a = \frac{y}{x^2}$. Substituting this into the derivative,we get $\frac{dy}{dx} = 2 \left(\frac{y}{x^2}\right) x = \frac{2y}{x}$.
The differential equation of the orthogonal trajectory is obtained by replacing $\frac{dy}{dx}$ with $-\frac{dx}{dy}$:
$-\frac{dx}{dy} = \frac{2y}{x}$.
Rearranging the terms,we get $x dx = -2y dy$,or $x dx + 2y dy = 0$.
Integrating both sides,we get $\int x dx + \int 2y dy = C$,which gives $\frac{x^2}{2} + y^2 = C$.

(D) Let the curve be $y = f(x)$. The area $A$ bounded by the curve,the $x$-axis,and the ordinate at $x$ is given by $A = \int_0^x y \, dx$.
The arc length $s$ of the curve from $x=0$ to $x$ is given by $s = \int_0^x \sqrt{1 + (y')^2} \, dx$.
According to the problem,$A = s$,so $\int_0^x y \, dx = \int_0^x \sqrt{1 + (y')^2} \, dx$.
Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus,we get $y = \sqrt{1 + (y')^2}$.
Squaring both sides,$y^2 = 1 + (y')^2$,which implies $(y')^2 = y^2 - 1$.
Thus,$y' = \pm \sqrt{y^2 - 1}$.
Separating variables,$\frac{dy}{\sqrt{y^2 - 1}} = \pm dx$.
Integrating both sides,$\ln|y + \sqrt{y^2 - 1}| = \pm x + C$.
Given the curve passes through $(0, 1)$,at $x=0$,$y=1$.
Substituting these values: $\ln|1 + \sqrt{1-1}| = \pm 0 + C \implies \ln(1) = C \implies C = 0$.
So,$\ln|y + \sqrt{y^2 - 1}| = x$ or $-x$.
This simplifies to $y + \sqrt{y^2 - 1} = e^x$ or $y + \sqrt{y^2 - 1} = e^{-x}$.
Solving for $y$,we get $y = \frac{e^x + e^{-x}}{2} = \cosh(x)$.
Also,if $y=1$,then $y'=0$,which satisfies $y = \sqrt{1+(y')^2} \implies 1 = \sqrt{1+0} = 1$.
Thus,both $y = \cosh(x)$ and $y=1$ are solutions.
Therefore,the correct option is $(D)$.

(C) The given differential equation is $x dy + y dx = 0$,which can be written as $d(xy) = 0$.
Integrating both sides,we get $xy = c$.
Since the curve passes through $(2, 8)$,we have $2 \times 8 = c$,so $c = 16$.
The equation of the conic is $xy = 16$,which represents a rectangular hyperbola.
For a rectangular hyperbola $xy = c$,the equation can be transformed to the standard form $X^2 - Y^2 = a^2$ by rotating the axes by $45^\circ$.
Here,$xy = 16$ is equivalent to $\frac{x^2}{32} - \frac{y^2}{32} = -1$ (or $Y^2 - X^2 = 32$ in rotated coordinates).
Comparing with $Y^2 - X^2 = a^2$,we have $a^2 = 32$,so $a = 4\sqrt{2}$.
The length of the latus rectum for a rectangular hyperbola $xy = c$ is $2\sqrt{2} \times \text{semi-transverse axis} \times \sqrt{2} = 4a$ is not correct; rather,for $xy = c$,the latus rectum is $2\sqrt{2} \times \sqrt{2c} = 4\sqrt{c}$.
Alternatively,for $xy = 16$,$a = 4\sqrt{2}$. The latus rectum is $2a = 2(4\sqrt{2}) = 8\sqrt{2}$.

(C) The rate of change of the water level $y$ is given by the differential equation: $\frac{dy}{dt} = -k \sqrt{y}$.
Separating the variables,we get: $\frac{dy}{\sqrt{y}} = -k \, dt$.
Integrating both sides from the initial depth $y = 4$ to the final depth $y = 0$ over the time interval from $t = 0$ to $t = T$:
$\int_{4}^{0} y^{-1/2} \, dy = \int_{0}^{T} -k \, dt$.
Evaluating the integrals:
$[2\sqrt{y}]_{4}^{0} = -k[t]_{0}^{T}$.
Substituting the limits:
$2(\sqrt{0} - \sqrt{4}) = -k(T - 0)$.
$2(0 - 2) = -kT$.
$-4 = -kT$.
$T = \frac{4}{k}$.
Given $k = \frac{1}{15}$,we have:
$T = \frac{4}{1/15} = 4 \times 15 = 60 \text{ min}$.

(A) Let the point of contact be $P(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.
The equation of the tangent at $P(x, y)$ is $Y - y = \frac{dy}{dx}(X - x)$.
To find the $x-$intercept,set $Y = 0$: $-y = \frac{dy}{dx}(X - x) \Rightarrow X - x = -y \frac{dx}{dy} \Rightarrow X = x - y \frac{dx}{dy}$.
According to the problem,the $x-$intercept is equal to the ordinate $y$,so $x - y \frac{dx}{dy} = y$.
Rearranging the terms: $x - y = y \frac{dx}{dy} \Rightarrow (x - y) dy = y dx \Rightarrow x dy - y dy = y dx$.
This can be written as $x dy - y dx = y dy$.
Dividing both sides by $y^2$: $\frac{x dy - y dx}{y^2} = \frac{y dy}{y^2} = \frac{dy}{y}$.
This is the derivative of the quotient: $d(\frac{x}{y}) = \frac{dy}{y}$.
Integrating both sides: $\frac{x}{y} = \ln|y| + C$.
Since the curve passes through $(1, 1)$,we substitute $x=1, y=1$: $\frac{1}{1} = \ln(1) + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$.
Thus,the equation is $\frac{x}{y} = \ln y + 1 \Rightarrow \frac{x}{y} - 1 = \ln y \Rightarrow \ln y = \frac{x-y}{y}$.
Exponentiating both sides: $y = e^{\frac{x-y}{y}} = e^{\frac{x}{y} - 1} = e^{\frac{x}{y}} \cdot e^{-1}$.
Therefore,$y e = e^{\frac{x}{y}} \Rightarrow y e^{1 - \frac{x}{y}} = \dots$ wait,let's re-evaluate: $y = e^{\frac{x}{y}} \cdot e^{-1} \Rightarrow y e = e^{\frac{x}{y}} \Rightarrow y e^{-\frac{x}{y}} = e^{-1}$.
Actually,from $\frac{x}{y} = \ln y + 1$,we have $\ln y = \frac{x}{y} - 1$,so $y = e^{\frac{x}{y} - 1} = \frac{e^{x/y}}{e}$.
Thus,$y e = e^{x/y}$,which is $y e^{-\frac{x}{y}} = e^{-1}$ or $y e^{1 - x/y} = 1$.
Checking the options,$y e^{x/y} = e$ is the standard form derived from $y = e^{x/y - 1} \cdot e = e^{x/y}$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{ax + b}{cy + d}$.
By separating the variables,we get: $(cy + d) dy = (ax + b) dx$.
Integrating both sides: $\int (cy + d) dy = \int (ax + b) dx$.
This yields: $\frac{c y^2}{2} + dy = \frac{a x^2}{2} + bx + K$,where $K$ is the constant of integration.
Rearranging the terms: $c y^2 - a x^2 + 2dy - 2bx - 2K = 0$.
For this equation to represent a parabola,one of the squared terms ($y^2$ or $x^2$) must be present,but not both with the same sign and magnitude (which would represent a circle or ellipse).
Specifically,if $c = 0$ and $a \neq 0$,the equation becomes $-a x^2 + 2dy - 2bx - 2K = 0$,which is a parabola opening along the $x$-axis.
If $a = 0$ and $c \neq 0$,the equation becomes $c y^2 + 2dy - 2bx - 2K = 0$,which is a parabola opening along the $y$-axis.
Thus,the condition is $a = 0, c \neq 0$ or $c = 0, a \neq 0$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{1+x}{2y}$.
Separating the variables,we get: $2y \, dy = (1+x) \, dx$.
Integrating both sides: $\int 2y \, dy = \int (1+x) \, dx$.
$y^2 = x + \frac{x^2}{2} + C$.
Since the conic passes through $(1, 1)$,substitute $x=1$ and $y=1$: $1^2 = 1 + \frac{1^2}{2} + C \Rightarrow 1 = 1.5 + C \Rightarrow C = -0.5$.
So,$y^2 = x + \frac{x^2}{2} - 0.5$.
Rearranging: $\frac{x^2}{2} + x - y^2 = 0.5$.
Multiply by $2$: $x^2 + 2x - 2y^2 = 1$.
Complete the square for $x$: $(x^2 + 2x + 1) - 2y^2 = 1 + 1$.
$(x+1)^2 - 2y^2 = 2$.
Divide by $2$: $\frac{(x+1)^2}{2} - \frac{y^2}{1} = 1$.
This is a hyperbola of the form $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$ where $a^2 = 2$ and $b^2 = 1$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$.

(B) Given differential equation: $(9 - x^2)(\frac{dy}{dx})^2 = 9 - y^2$
$\Rightarrow (\frac{dy}{dx})^2 = \frac{9 - y^2}{9 - x^2}$
$\Rightarrow \frac{dy}{dx} = \pm \frac{\sqrt{9 - y^2}}{\sqrt{9 - x^2}}$
Separating the variables:
$\frac{dy}{\sqrt{9 - y^2}} = \pm \frac{dx}{\sqrt{9 - x^2}}$
Integrating both sides:
$\int \frac{dy}{\sqrt{3^2 - y^2}} = \pm \int \frac{dx}{\sqrt{3^2 - x^2}}$
$\sin^{-1}(\frac{y}{3}) = \pm \sin^{-1}(\frac{x}{3}) + C$
Since the curve passes through $(3, 0)$:
$\sin^{-1}(0) = \pm \sin^{-1}(\frac{3}{3}) + C$
$0 = \pm \frac{\pi}{2} + C \Rightarrow C = \mp \frac{\pi}{2}$
Substituting $C$ back:
$\sin^{-1}(\frac{y}{3}) = \pm \sin^{-1}(\frac{x}{3}) \mp \frac{\pi}{2}$
Taking $\sin$ on both sides:
$\frac{y}{3} = \sin(\pm \sin^{-1}(\frac{x}{3}) \mp \frac{\pi}{2}) = \mp \cos(\sin^{-1}(\frac{x}{3})) = \mp \sqrt{1 - \frac{x^2}{9}}$
Squaring both sides:
$\frac{y^2}{9} = 1 - \frac{x^2}{9} \Rightarrow x^2 + y^2 = 9$
This is the equation of a circle with radius $3$ centered at the origin.

(A) The length of the subnormal is given by $y \frac{dy}{dx}$.
Given that the length of the subnormal is one more than the square of the ordinate $(y^2)$,we have the differential equation:
$y \frac{dy}{dx} = y^2 + 1$
Rearranging the terms,we get:
$\frac{y}{y^2 + 1} dy = dx$
Integrating both sides:
$\int \frac{y}{y^2 + 1} dy = \int dx$
$\frac{1}{2} \ln(y^2 + 1) = x + C$
Since the curve passes through the origin $(0, 0)$,we substitute $x=0$ and $y=0$:
$\frac{1}{2} \ln(0^2 + 1) = 0 + C \implies C = 0$
Thus,$\frac{1}{2} \ln(y^2 + 1) = x$
$\ln(y^2 + 1) = 2x$
$y^2 + 1 = e^{2x}$
$y^2 = e^{2x} - 1$
$y = \sqrt{e^{2x} - 1}$
Therefore,the correct option is $A$.

(C) Given the differential equation: $2xy \, dy = (x^2 + y^2 + 1) dx$
Rearranging the terms: $2xy \, dy - y^2 dx = (x^2 + 1) dx$
Divide both sides by $x^2$: $\frac{2xy \, dy - y^2 dx}{x^2} = \frac{(x^2 + 1) dx}{x^2}$
This can be written as: $d\left(\frac{y^2}{x}\right) = (1 + x^{-2}) dx$
Integrating both sides: $\frac{y^2}{x} = x - \frac{1}{x} + C$
Multiplying by $x$: $y^2 = x^2 - 1 + Cx$
Rearranging to standard form: $x^2 - Cx - y^2 = 1$
Completing the square for $x$: $(x - \frac{C}{2})^2 - y^2 = 1 + \frac{C^2}{4}$
This equation represents a family of rectangular hyperbolas with the center at $(\frac{C}{2}, 0)$,which lies on the $x$-axis.

(C) Let the tangent at $P(x, y)$ be $Y - y = f'(x)(X - x)$.
For $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$. So $A = \left( x - \frac{y}{f'(x)}, 0 \right)$.
For $B$ (where $X=0$),$Y = y - x f'(x)$. So $B = (0, y - x f'(x))$.
Given $AP : BP = 1 : 3$,by section formula for $P(x, y)$ dividing $AB$ in ratio $1:3$:
$x = \frac{1 \cdot 0 + 3 \cdot (x - y/f'(x))}{1 + 3} \implies 4x = 3x - \frac{3y}{f'(x)} \implies x = -\frac{3y}{f'(x)}$.
Thus,$\frac{dy}{dx} = -\frac{3y}{x}$.
Separating variables: $\int \frac{dy}{y} = -3 \int \frac{dx}{x} \implies \ln|y| = -3 \ln|x| + C \implies y = \frac{k}{x^3}$.
Given $f(1) = 1$,we get $1 = \frac{k}{1^3} \implies k = 1$. So $y = \frac{1}{x^3}$.
Checking the options,for $x=2$,$y = \frac{1}{2^3} = \frac{1}{8}$.
Thus,the curve passes through $\left( 2, \frac{1}{8} \right)$.

(D) Given the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$.
Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting this into the differential equation,we get $v + x \frac{dv}{dx} = v + \phi \left( \frac{1}{v} \right)$,which simplifies to $x \frac{dv}{dx} = \phi \left( \frac{1}{v} \right)$.
Rearranging the terms,we have $\frac{dv}{\phi(1/v)} = \frac{dx}{x}$.
Integrating both sides,$\int \frac{dv}{\phi(1/v)} = \ln |x| + C_1$.
Given the general solution $y \ln |cx| = x$,we can rewrite this as $\ln |cx| = \frac{x}{y} = \frac{1}{v}$.
Thus,$\ln |x| + \ln |c| = \frac{1}{v}$.
Differentiating both sides with respect to $v$,we get $\frac{1}{x} \frac{dx}{dv} = -\frac{1}{v^2}$.
From our earlier equation $x \frac{dv}{dx} = \phi(1/v)$,we have $\frac{dx}{dv} = \frac{x}{\phi(1/v)}$.
Substituting this into the derivative result: $\frac{1}{x} \cdot \frac{x}{\phi(1/v)} = -\frac{1}{v^2}$,which implies $\phi(1/v) = -v^2$.
We want to find $\phi(2)$. Let $\frac{1}{v} = 2$,so $v = \frac{1}{2}$.
Then $\phi(2) = -(\frac{1}{2})^2 = -\frac{1}{4}$.

(B) Given the functional equation $f(xy) = f(x)f(y)$ for all $x, y \in [0, 1].$
Setting $y = 0,$ we get $f(0) = f(x)f(0).$
Since $f(0) \ne 0,$ we can divide by $f(0)$ to obtain $f(x) = 1$ for all $x \in [0, 1].$
Now,the differential equation is $\frac{dy}{dx} = f(x) = 1.$
Integrating both sides with respect to $x,$ we get $y = x + c.$
Using the initial condition $y(0) = 1,$ we find $1 = 0 + c,$ so $c = 1.$
Thus,the function is $y(x) = x + 1.$
We need to calculate $y\left( \frac{1}{4} \right) + y\left( \frac{3}{4} \right).$
$y\left( \frac{1}{4} \right) = \frac{1}{4} + 1 = \frac{5}{4}.$
$y\left( \frac{3}{4} \right) = \frac{3}{4} + 1 = \frac{7}{4}.$
Therefore,$y\left( \frac{1}{4} \right) + y\left( \frac{3}{4} \right) = \frac{5}{4} + \frac{7}{4} = \frac{12}{4} = 3.$

(A) Given function: $y = \sqrt{a^{2} - x^{2}}$
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{a^{2} - x^{2}})$
Using the chain rule:
$\frac{dy}{dx} = \frac{1}{2\sqrt{a^{2} - x^{2}}} \cdot \frac{d}{dx}(a^{2} - x^{2})$
$\frac{dy}{dx} = \frac{1}{2\sqrt{a^{2} - x^{2}}} \cdot (-2x)$
$\frac{dy}{dx} = \frac{-x}{\sqrt{a^{2} - x^{2}}}$
Now,substitute the value of $\frac{dy}{dx}$ into the given differential equation $x + y \frac{dy}{dx} = 0$:
$L.H.S = x + y \left( \frac{-x}{\sqrt{a^{2} - x^{2}}} \right)$
Since $y = \sqrt{a^{2} - x^{2}}$,we substitute this into the expression:
$L.H.S = x + \sqrt{a^{2} - x^{2}} \cdot \left( \frac{-x}{\sqrt{a^{2} - x^{2}}} \right)$
$L.H.S = x - x = 0$
$L.H.S = R.H.S$
Hence,the given function is a solution of the differential equation.

(A) Let $P$ be the principal at any time $t$. According to the given problem,the rate of increase is $\frac{dP}{dt} = \frac{5}{100} P = \frac{P}{20}$.
Separating the variables,we get $\frac{dP}{P} = \frac{dt}{20}$.
Integrating both sides,we get $\int \frac{dP}{P} = \int \frac{dt}{20}$,which gives $\log_e P = \frac{t}{20} + C_1$.
This implies $P = e^{\frac{t}{20} + C_1} = C e^{\frac{t}{20}}$,where $C = e^{C_1}$.
At $t = 0$,$P = 1000$. Substituting these values,$1000 = C e^0$,so $C = 1000$.
Thus,the equation is $P = 1000 e^{\frac{t}{20}}$.
To find the time $t$ when the principal doubles,we set $P = 2000$:
$2000 = 1000 e^{\frac{t}{20}}$
$2 = e^{\frac{t}{20}}$
Taking the natural logarithm on both sides,$\log_e 2 = \frac{t}{20}$.
Therefore,$t = 20 \log_e 2$ years.

(A) The given differential equation is $y^{\prime} = e^{x} \sin x$.
This can be written as $\frac{dy}{dx} = e^{x} \sin x$.
Integrating both sides with respect to $x$:
$y = \int e^{x} \sin x \, dx + C$ ...........$(1)$
To evaluate $I = \int e^{x} \sin x \, dx$,we use integration by parts:
$I = \sin x \cdot e^{x} - \int \cos x \cdot e^{x} \, dx$
$I = e^{x} \sin x - [\cos x \cdot e^{x} - \int(-\sin x) \cdot e^{x} \, dx]$
$I = e^{x} \sin x - e^{x} \cos x - I$
$2I = e^{x}(\sin x - \cos x)$
$I = \frac{e^{x}(\sin x - \cos x)}{2}$
Substituting this into equation $(1)$:
$y = \frac{e^{x}(\sin x - \cos x)}{2} + C$ ...........$(2)$
Since the curve passes through $(0,0)$,we substitute $x=0$ and $y=0$:
$0 = \frac{e^{0}(\sin 0 - \cos 0)}{2} + C$
$0 = \frac{1(0 - 1)}{2} + C$
$0 = -\frac{1}{2} + C \Rightarrow C = \frac{1}{2}$
Substituting $C = \frac{1}{2}$ into equation $(2)$:
$y = \frac{e^{x}(\sin x - \cos x)}{2} + \frac{1}{2}$
$2y = e^{x}(\sin x - \cos x) + 1$
$2y - 1 = e^{x}(\sin x - \cos x)$.

(A) Let $(x, y)$ be any point on the curve.
The slope $(m_1)$ of the line segment joining $(x, y)$ and $(-4, -3)$ is given by $m_1 = \frac{y - (-3)}{x - (-4)} = \frac{y+3}{x+4}$.
The slope $(m_2)$ of the tangent to the curve at $(x, y)$ is $\frac{dy}{dx}$.
According to the problem,$m_2 = 2m_1$.
Therefore,$\frac{dy}{dx} = 2 \left( \frac{y+3}{x+4} \right)$.
Separating the variables,we get $\frac{dy}{y+3} = \frac{2 dx}{x+4}$.
Integrating both sides,$\int \frac{dy}{y+3} = 2 \int \frac{dx}{x+4}$.
This gives $\ln|y+3| = 2 \ln|x+4| + C_1$,which can be written as $\ln|y+3| = \ln|x+4|^2 + \ln C$.
Thus,$y+3 = C(x+4)^2$.
Since the curve passes through $(-2, 1)$,we substitute $x = -2$ and $y = 1$:
$1+3 = C(-2+4)^2 \Rightarrow 4 = C(2)^2 \Rightarrow 4 = 4C \Rightarrow C = 1$.
Substituting $C=1$ into the general equation,we get $y+3 = (x+4)^2$.

(A) Let the volume of the spherical balloon be $V = \frac{4}{3} \pi r^3$.
Given that the rate of change of volume is constant,let $\frac{dV}{dt} = k$,where $k$ is a constant.
Integrating with respect to $t$,we get $V = kt + C$.
Substituting $V = \frac{4}{3} \pi r^3$,we have $\frac{4}{3} \pi r^3 = kt + C$.
At $t = 0$,$r = 3$,so $\frac{4}{3} \pi (3)^3 = k(0) + C \Rightarrow C = 36 \pi$.
At $t = 3$,$r = 6$,so $\frac{4}{3} \pi (6)^3 = k(3) + 36 \pi$.
$288 \pi = 3k + 36 \pi \Rightarrow 3k = 252 \pi \Rightarrow k = 84 \pi$.
Substituting $k$ and $C$ back into the equation: $\frac{4}{3} \pi r^3 = 84 \pi t + 36 \pi$.
Dividing by $\frac{4}{3} \pi$,we get $r^3 = 63t + 27$.
Therefore,$r = (63t + 27)^{\frac{1}{3}}$.

(A) Let $P$ be the principal at time $t$. It is given that the principal increases continuously at the rate of $r \%$ per year.
$\frac{dP}{dt} = \left(\frac{r}{100}\right) P$
Separating the variables,we get:
$\frac{dP}{P} = \frac{r}{100} dt$
Integrating both sides:
$\int \frac{dP}{P} = \int \frac{r}{100} dt$
$\log_{e} P = \frac{rt}{100} + C$
At $t = 0$,$P = 100$. Substituting these values:
$\log_{e} 100 = \frac{r(0)}{100} + C \Rightarrow C = \log_{e} 100$
So,$\log_{e} P = \frac{rt}{100} + \log_{e} 100$
$\log_{e} P - \log_{e} 100 = \frac{rt}{100}$
$\log_{e} \left(\frac{P}{100}\right) = \frac{rt}{100}$
Given that the principal doubles in $10$ years,so at $t = 10$,$P = 200$:
$\log_{e} \left(\frac{200}{100}\right) = \frac{r(10)}{100}$
$\log_{e} 2 = \frac{r}{10}$
Given $\log_{e} 2 = 0.6931$:
$0.6931 = \frac{r}{10}$
$r = 6.931 \%$
Thus,the value of $r$ is $6.931 \%$.

(A) Let $P$ be the principal at time $t$.
Given that the principal increases continuously at the rate of $5 \%$ per year,we have the differential equation:
$\frac{dP}{dt} = \frac{5}{100} P = 0.05 P$
Separating the variables,we get:
$\frac{dP}{P} = 0.05 dt$
Integrating both sides:
$\int \frac{dP}{P} = \int 0.05 dt$
$\ln P = 0.05 t + C$
$P = e^{0.05 t + C} = e^C \cdot e^{0.05 t}$
Let $P_0 = e^C$ be the initial principal at $t = 0$. Given $P_0 = 1000$,so $P = 1000 e^{0.05 t}$.
For $t = 10$ years:
$P = 1000 e^{0.05 \times 10} = 1000 e^{0.5}$
Using the given value $e^{0.5} = 1.648$:
$P = 1000 \times 1.648 = 1648$
Thus,the amount after $10$ years will be $Rs. 1648$.

(A) Let $y$ be the number of bacteria at any time $t$.
Given that the rate of growth is proportional to the number present,we have $\frac{dy}{dt} = ky$,where $k$ is a constant.
Separating variables and integrating,we get $\int \frac{dy}{y} = \int k dt$,which gives $\log y = kt + C$.
At $t = 0$,let $y = y_0 = 1,00,000$. Thus,$C = \log y_0$.
So,$\log y = kt + \log y_0$,or $\log \left(\frac{y}{y_0}\right) = kt$.
Given that the number increases by $10 \%$ in $2$ hours,at $t = 2$,$y = y_0 + 0.1 y_0 = 1.1 y_0$.
Substituting this into the equation: $\log(1.1) = k(2) \Rightarrow k = \frac{1}{2} \log(1.1)$.
We want to find $t$ when $y = 2,00,000 = 2 y_0$.
Substituting $y = 2 y_0$ and $k = \frac{1}{2} \log(1.1)$ into $\log \left(\frac{y}{y_0}\right) = kt$:
$\log(2) = \left(\frac{1}{2} \log(1.1)\right) t$.
Solving for $t$,we get $t = \frac{2 \log 2}{\log(1.1)}$ hours.

The given function is $y=e^{a x}\left[c_{1} \cos b x+c_{2} \sin b x\right]$ .........$(1)$
Differentiating both sides of equation $(1)$ with respect to $x,$ we get
$\frac{d y}{d x}=e^{a x}\left[-b c_{1} \sin b x+b c_{2} \cos b x\right]+\left[c_{1} \cos b x+c_{2} \sin b x\right] e^{a x} \cdot a$
$\frac{d y}{d x}=e^{a x}\left[\left(b c_{2}+a c_{1}\right) \cos b x+\left(a c_{2}-b c_{1}\right) \sin b x\right]$ .........$(2)$
Differentiating both sides of equation $(2)$ with respect to $x,$ we get
$\frac{d^{2} y}{d x^{2}}=e^{a x}\left[\left(b c_{2}+a c_{1}\right)(-b \sin b x)+\left(a c_{2}-b c_{1}\right)(b \cos b x)\right]+\left[\left(b c_{2}+a c_{1}\right) \cos b x+\left(a c_{2}-b c_{1}\right) \sin b x\right] e^{a x} \cdot a$
$=e^{a x}\left[\left(a^{2} c_{2}-2 a b c_{1}-b^{2} c_{2}\right) \sin b x+\left(a^{2} c_{1}+2 a b c_{2}-b^{2} c_{1}\right) \cos b x\right]$
Substituting the values of $\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}$ and $y$ in the given differential equation,we get
$L.H.S. = e^{a x}\left[\left(a^{2} c_{2}-2 a b c_{1}-b^{2} c_{2}\right) \sin b x+\left(a^{2} c_{1}+2 a b c_{2}-b^{2} c_{1}\right) \cos b x\right] - 2 a e^{a x}\left[\left(b c_{2}+a c_{1}\right) \cos b x+\left(a c_{2}-b c_{1}\right) \sin b x\right] + \left(a^{2}+b^{2}\right) e^{a x}\left[c_{1} \cos b x+c_{2} \sin b x\right]$
$=e^{a x}\left[\left(a^{2} c_{2}-2 a b c_{1}-b^{2} c_{2}-2 a^{2} c_{2}+2 a b c_{1}+a^{2} c_{2}+b^{2} c_{2}\right) \sin b x + \left(a^{2} c_{1}+2 a b c_{2}-b^{2} c_{1}-2 a b c_{2}-2 a^{2} c_{1}+a^{2} c_{1}+b^{2} c_{1}\right) \cos b x\right]$
$=e^{a x}[0 \cdot \sin b x + 0 \cdot \cos b x] = 0 = R.H.S.$
Hence,the given function is a solution of the given differential equation.

(A) Given function: $y = x \sin 3x$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \sin 3x + x \cdot \frac{d}{dx}(\sin 3x)$
$\frac{dy}{dx} = 1 \cdot \sin 3x + x \cdot (\cos 3x \cdot 3) = \sin 3x + 3x \cos 3x$
Differentiating again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\sin 3x) + 3 \cdot \frac{d}{dx}(x \cos 3x)$
$\frac{d^{2}y}{dx^{2}} = 3 \cos 3x + 3 [1 \cdot \cos 3x + x \cdot (-\sin 3x \cdot 3)]$
$\frac{d^{2}y}{dx^{2}} = 3 \cos 3x + 3 \cos 3x - 9x \sin 3x = 6 \cos 3x - 9x \sin 3x$
Now,substitute $\frac{d^{2}y}{dx^{2}}$ and $y$ into the $L.H.S.$ of the differential equation:
$L.H.S. = \frac{d^{2}y}{dx^{2}} + 9y - 6 \cos 3x$
$= (6 \cos 3x - 9x \sin 3x) + 9(x \sin 3x) - 6 \cos 3x$
$= 6 \cos 3x - 6 \cos 3x - 9x \sin 3x + 9x \sin 3x$
$= 0 = R.H.S.$
Since $L.H.S. = R.H.S.$,the given function is a solution of the differential equation.

(A) Let the population at any time $t$ be $y$.
It is given that the rate of increase of population is proportional to the number of inhabitants at any time $t$.
$\frac{dy}{dt} = ky$ (where $k$ is a constant)
$\frac{dy}{y} = k dt$
Integrating both sides,we get:
$\ln y = kt + C$ ............$(1)$
In the year $1999$,let $t = 0$ and $y = 20,000$.
$\ln(20,000) = k(0) + C \Rightarrow C = \ln(20,000)$.
In the year $2004$,$t = 5$ and $y = 25,000$.
$\ln(25,000) = 5k + \ln(20,000)$
$5k = \ln\left(\frac{25,000}{20,000}\right) = \ln\left(\frac{5}{4}\right)$
$k = \frac{1}{5} \ln\left(\frac{5}{4}\right)$.
In the year $2009$,$t = 10$.
Substituting $t = 10, k = \frac{1}{5} \ln\left(\frac{5}{4}\right)$,and $C = \ln(20,000)$ in equation $(1)$:
$\ln y = 10 \times \frac{1}{5} \ln\left(\frac{5}{4}\right) + \ln(20,000)$
$\ln y = 2 \ln\left(\frac{5}{4}\right) + \ln(20,000)$
$\ln y = \ln\left(\left(\frac{5}{4}\right)^2 \times 20,000\right)$
$y = 20,000 \times \frac{25}{16} = 1,250 \times 25 = 31,250$.
Thus,the population in $2009$ will be $31,250$.

(A) Given the differential equation $\frac{dy}{dx} = 2(x + 1)$.
Integrating both sides with respect to $x$,we get:
$y(x) = \int 2(x + 1) dx = (x + 1)^2 + C = x^2 + 2x + 1 + C$.
Let $K = 1 + C$,so $y(x) = (x + 1)^2 + C$.
The curve is a parabola opening upwards with vertex at $(-1, C)$. For the curve to bound an area with the $x-$axis,it must lie below the $x-$axis,so $C < 0$. Let $C = -k^2$ where $k > 0$.
The roots of $y(x) = 0$ are $(x + 1)^2 = -C$,so $x = -1 \pm \sqrt{-C}$.
The area bounded by the curve and the $x-$axis is given by:
$A = \int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} (0 - ((x + 1)^2 + C)) dx = -\int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} ((x + 1)^2 + C) dx$.
Let $u = x + 1$,then $du = dx$. The limits change to $-\sqrt{-C}$ to $\sqrt{-C}$.
$A = -\int_{-\sqrt{-C}}^{\sqrt{-C}} (u^2 + C) du = -[\frac{u^3}{3} + Cu]_{-\sqrt{-C}}^{\sqrt{-C}} = -[(\frac{(-C)^{3/2}}{3} + C\sqrt{-C}) - (\frac{-(-C)^{3/2}}{3} - C\sqrt{-C})]$.
$A = -[\frac{2}{3}(-C)^{3/2} + 2C\sqrt{-C}] = -[\frac{2}{3}(-C)\sqrt{-C} - 2(-C)\sqrt{-C}] = -[-\frac{4}{3}(-C)^{3/2}] = \frac{4}{3}(-C)^{3/2}$.
Given $A = \frac{4\sqrt{8}}{3} = \frac{4(2\sqrt{2})}{3} = \frac{8\sqrt{2}}{3}$.
So,$\frac{4}{3}(-C)^{3/2} = \frac{8\sqrt{2}}{3} \Rightarrow (-C)^{3/2} = 2\sqrt{2} = (\sqrt{2})^3$.
Thus,$-C = 2$,which means $C = -2$.
The function is $y(x) = (x + 1)^2 - 2 = x^2 + 2x - 1$.
Therefore,$y(1) = (1)^2 + 2(1) - 1 = 1 + 2 - 1 = 2$.

(C) Given the differential equation: $x dy - y dx = \sqrt{x^2 - y^2} dx$.
Dividing by $x^2$ (for $x \geq 1$): $\frac{x dy - y dx}{x^2} = \frac{1}{x} \sqrt{1 - (\frac{y}{x})^2} dx$.
This simplifies to: $d(\frac{y}{x}) = \frac{1}{x} \sqrt{1 - (\frac{y}{x})^2} dx$.
Integrating both sides: $\int \frac{d(\frac{y}{x})}{\sqrt{1 - (\frac{y}{x})^2}} = \int \frac{dx}{x}$.
$\sin^{-1}(\frac{y}{x}) = \ln|x| + C$.
Using the initial condition $y(1) = 0$: $\sin^{-1}(0) = \ln(1) + C \Rightarrow C = 0$.
Thus,$y = x \sin(\ln x)$.
The area $A$ is given by: $A = \int_{1}^{e^{\pi}} x \sin(\ln x) dx$.
Let $x = e^t$,then $dx = e^t dt$. When $x=1, t=0$; when $x=e^{\pi}, t=\pi$.
$A = \int_{0}^{\pi} e^t \sin(t) e^t dt = \int_{0}^{\pi} e^{2t} \sin(t) dt$.
Using the formula $\int e^{at} \sin(bt) dt = \frac{e^{at}}{a^2 + b^2} (a \sin(bt) - b \cos(bt)) + C$:
$A = [\frac{e^{2t}}{5} (2 \sin t - \cos t)]_{0}^{\pi} = \frac{e^{2\pi}}{5} (2(0) - (-1)) - \frac{1}{5} (2(0) - 1) = \frac{e^{2\pi}}{5} + \frac{1}{5}$.
Comparing with $\alpha e^{2\pi} + \beta$,we get $\alpha = \frac{1}{5}$ and $\beta = \frac{1}{5}$.
Therefore,$10(\alpha + \beta) = 10(\frac{1}{5} + \frac{1}{5}) = 10(\frac{2}{5}) = 4$.

(D) Given the differential equation: $\frac{dP}{dt} = 0.5P - 450 = 0.5(P - 900)$.
Separating the variables,we get: $\frac{dP}{P - 900} = 0.5 dt$.
Integrating both sides: $\int \frac{dP}{P - 900} = \int 0.5 dt$.
This gives: $\ln|P - 900| = 0.5t + C$.
Using the initial condition $P(0) = 850$: $\ln|850 - 900| = 0.5(0) + C \Rightarrow C = \ln(50)$.
So,the equation becomes: $\ln|P(t) - 900| = 0.5t + \ln(50)$.
We want to find $t$ when $P(t) = 0$: $\ln|0 - 900| = 0.5t + \ln(50)$.
$\ln(900) - \ln(50) = 0.5t$.
$\ln\left(\frac{900}{50}\right) = 0.5t$.
$\ln(18) = 0.5t$.
$t = 2 \ln(18)$.