Application of differential equations Questions in English

(A) Let $B(t)$ be the number of bacteria at time $t$. The rate of growth is given by $\frac{dB}{dt} = \lambda B$.
Integrating this,we get $B(t) = B_0 e^{\lambda t}$,where $B_0 = 1000$.
Given that at $t = 2$,$B(2) = 1000 + 20\% \text{ of } 1000 = 1200$.
So,$1200 = 1000 e^{2\lambda} \Rightarrow e^{2\lambda} = \frac{6}{5} \Rightarrow 2\lambda = \log_{e}\left(\frac{6}{5}\right) \Rightarrow \lambda = \frac{1}{2} \log_{e}\left(\frac{6}{5}\right)$.
We are given $B(T) = 2000$ where $T = \frac{k}{\log_{e}\left(\frac{6}{5}\right)}$.
Using $B(T) = B_0 e^{\lambda T}$,we have $2000 = 1000 e^{\lambda T} \Rightarrow 2 = e^{\lambda T} \Rightarrow \log_{e} 2 = \lambda T$.
Substituting $\lambda$ and $T$: $\log_{e} 2 = \left(\frac{1}{2} \log_{e}\left(\frac{6}{5}\right)\right) \times \left(\frac{k}{\log_{e}\left(\frac{6}{5}\right)}\right) = \frac{k}{2}$.
Thus,$k = 2 \log_{e} 2$.
Finally,$\left(\frac{k}{\log_{e} 2}\right)^{2} = \left(\frac{2 \log_{e} 2}{\log_{e} 2}\right)^{2} = 2^{2} = 4$.

(B) Given the slope of the tangent $\frac{dy}{dx} = \frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$.
Multiply numerator and denominator by $e^{2x}$:
$\frac{dy}{dx} = \frac{2e^{4x} - 6e^x + 9e^{2x}}{2e^{2x} + 9}$.
This simplifies to $\frac{dy}{dx} = e^{2x} - \frac{6e^x}{2e^{2x} + 9}$.
Integrating both sides with respect to $x$:
$y = \int e^{2x} dx - \int \frac{6e^x}{2e^{2x} + 9} dx$.
Let $u = \sqrt{2}e^x$,then $du = \sqrt{2}e^x dx$,so $e^x dx = \frac{du}{\sqrt{2}}$.
$y = \frac{1}{2}e^{2x} - \int \frac{6}{\sqrt{2}(u^2 + 3^2)} du = \frac{1}{2}e^{2x} - \frac{6}{\sqrt{2} \cdot 3} \tan^{-1}(\frac{u}{3}) + C$.
$y = \frac{1}{2}e^{2x} - \sqrt{2} \tan^{-1}(\frac{\sqrt{2}e^x}{3}) + C$.
Using the point $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$:
$\frac{1}{2} + \frac{\pi}{2\sqrt{2}} = \frac{1}{2} - \sqrt{2} \tan^{-1}(\frac{\sqrt{2}}{3}) + C \implies C = \frac{\pi}{2\sqrt{2}} + \sqrt{2} \tan^{-1}(\frac{\sqrt{2}}{3})$.
Using the point $(\alpha, \frac{1}{2}e^{2\alpha})$:
$\frac{1}{2}e^{2\alpha} = \frac{1}{2}e^{2\alpha} - \sqrt{2} \tan^{-1}(\frac{\sqrt{2}e^{\alpha}}{3}) + C$.
$\sqrt{2} \tan^{-1}(\frac{\sqrt{2}e^{\alpha}}{3}) = \frac{\pi}{2\sqrt{2}} + \sqrt{2} \tan^{-1}(\frac{\sqrt{2}}{3})$.
$\tan^{-1}(\frac{\sqrt{2}e^{\alpha}}{3}) = \frac{\pi}{4} + \tan^{-1}(\frac{\sqrt{2}}{3})$.
Taking $\tan$ on both sides:
$\frac{\sqrt{2}e^{\alpha}}{3} = \tan(\frac{\pi}{4} + \tan^{-1}(\frac{\sqrt{2}}{3})) = \frac{1 + \frac{\sqrt{2}}{3}}{1 - \frac{\sqrt{2}}{3}} = \frac{3+\sqrt{2}}{3-\sqrt{2}}$.
$e^{\alpha} = \frac{3}{\sqrt{2}} \left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$.

(A) Given the continuous function $f:[0, \infty) \rightarrow R$ satisfying $f(x) = 2 \int_0^x t f(t) d t + 1, \forall x \geq 0$.
Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$:
$f'(x) = 2x f(x)$.
Rearranging the terms,we get $\frac{f'(x)}{f(x)} = 2x$.
Integrating both sides with respect to $x$:
$\int \frac{f'(x)}{f(x)} dx = \int 2x dx$.
$\ln |f(x)| = x^2 + C$.
$f(x) = K e^{x^2}$,where $K = e^C$.
To find $K$,we evaluate the original equation at $x=0$:
$f(0) = 2 \int_0^0 t f(t) dt + 1 = 0 + 1 = 1$.
Substituting $x=0$ into $f(x) = K e^{x^2}$,we get $f(0) = K e^0 = K$.
Thus,$K = 1$,which gives $f(x) = e^{x^2}$.
Finally,$f(1) = e^{(1)^2} = e^1 = e$.

(D) Given the equation $\int_0^x f(t) dt = p f(x)$.
If $p = 0$,then $\int_0^x f(t) dt = 0$ for all $x \in \mathbb{R}$. Differentiating with respect to $x$,we get $f(x) = 0$ for all $x$. Thus,there is no non-zero function $f$ for $p = 0$.
If $p \neq 0$,we differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$f(x) = p f'(x) \implies \frac{f'(x)}{f(x)} = \frac{1}{p}$.
Integrating both sides,we get $\ln|f(x)| = \frac{x}{p} + C$,which implies $f(x) = A e^{x/p}$ for some constant $A$.
Substituting this back into the original integral equation at $x = 0$:
$\int_0^0 f(t) dt = p f(0) \implies 0 = p A e^0 \implies p A = 0$.
Since $p \neq 0$,we must have $A = 0$,which implies $f(x) = 0$ for all $x$.
Thus,for any $p \in \mathbb{R}$,there exists no non-zero continuous function $f$ satisfying the given equation.
Therefore,$S = \mathbb{R}$.

(D) Given the differential equation $\frac{dy}{dx} = 2 \sqrt{y}$ with the initial condition $y(0) = 0$.
Case $I$: The trivial solution $y(x) = 0$ for all $x \in \mathbb{R}$ satisfies the equation and the initial condition.
Case $II$: For any constant $a \geq 0$,we can define a family of functions:
$y(x) = \begin{cases} 0 & x < a \\ (x-a)^2 & x \geq a \end{cases}$
Let us check the differentiability at $x = a$:
Left-hand derivative: $\lim_{h \to 0^-} \frac{y(a+h) - y(a)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$.
Right-hand derivative: $\lim_{h \to 0^+} \frac{y(a+h) - y(a)}{h} = \lim_{h \to 0^+} \frac{(a+h-a)^2 - 0}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = 0$.
Since the left-hand derivative equals the right-hand derivative,the function is differentiable at $x = a$.
For $x > a$,$y'(x) = 2(x-a) = 2\sqrt{(x-a)^2} = 2\sqrt{y(x)}$.
Since $a$ can be any non-negative real number,there are infinitely many such functions.

(D) Given $f(x) + \int_0^x f(t) \sqrt{1 - (\ln f(t))^2} dt = e$.
At $x=0$,$f(0) + 0 = e$,so $f(0) = e$.
Differentiating both sides with respect to $x$ using Leibniz's rule:
$f'(x) + f(x) \sqrt{1 - (\ln f(x))^2} = 0$.
Let $y = f(x)$,then $\frac{dy}{dx} = -y \sqrt{1 - (\ln y)^2}$.
Separating variables: $\int \frac{dy}{y \sqrt{1 - (\ln y)^2}} = -\int dx$.
Let $\ln y = t$,then $\frac{1}{y} dy = dt$.
$\int \frac{dt}{\sqrt{1 - t^2}} = -x + C$.
$\sin^{-1}(t) = -x + C \Rightarrow \sin^{-1}(\ln f(x)) = -x + C$.
Since $f(0) = e$,$\sin^{-1}(\ln e) = -0 + C \Rightarrow \sin^{-1}(1) = C \Rightarrow C = \frac{\pi}{2}$.
Thus,$\sin^{-1}(\ln f(x)) = \frac{\pi}{2} - x$.
For $x = \frac{\pi}{6}$,$\sin^{-1}(\ln f(\frac{\pi}{6})) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.
Therefore,$\ln f(\frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.
Finally,$(6 \ln f(\frac{\pi}{6}))^2 = (6 \times \frac{\sqrt{3}}{2})^2 = (3\sqrt{3})^2 = 27$.

(D) Given the differential equation $\frac{dy}{dx} = -\frac{x+a}{y-2}$.
Integrating both sides,we get $\int (y-2) dy = -\int (x+a) dx$,which gives $\frac{(y-2)^2}{2} = -\frac{(x+a)^2}{2} + k$.
This simplifies to $(x+a)^2 + (y-2)^2 = 2k$. Since the area is $4\pi$,the radius $r = 2$,so $2k = 4$,meaning $(x+a)^2 + (y-2)^2 = 4$.
Using $y(1) = 0$,we have $(1+a)^2 + (0-2)^2 = 4$,so $(1+a)^2 = 0$,which gives $a = -1$.
The equation of the circle is $(x-1)^2 + (y-2)^2 = 4$.
For intersection with the $y$-axis,set $x=0$: $(0-1)^2 + (y-2)^2 = 4 \implies (y-2)^2 = 3 \implies y = 2 \pm \sqrt{3}$.
So $P = (0, 2+\sqrt{3})$ and $Q = (0, 2-\sqrt{3})$.
The center of the circle is $(1, 2)$. The slope of the radius to $P$ is $\frac{(2+\sqrt{3})-2}{0-1} = -\sqrt{3}$.
The slope of the normal at $P$ is the same as the slope of the radius,which is $-\sqrt{3}$.
The equation of the normal at $P$ is $y - (2+\sqrt{3}) = -\sqrt{3}(x-0) \implies y = -\sqrt{3}x + 2 + \sqrt{3}$.
Setting $y=0$ for $R$,we get $0 = -\sqrt{3}x + 2 + \sqrt{3} \implies x_R = 1 + \frac{2}{\sqrt{3}}$.
Similarly,for $Q$,the slope of the radius is $\frac{(2-\sqrt{3})-2}{0-1} = \sqrt{3}$.
The equation of the normal at $Q$ is $y - (2-\sqrt{3}) = \sqrt{3}(x-0) \implies y = \sqrt{3}x + 2 - \sqrt{3}$.
Setting $y=0$ for $S$,we get $0 = \sqrt{3}x + 2 - \sqrt{3} \implies x_S = 1 - \frac{2}{\sqrt{3}}$.
The length $RS = |x_R - x_S| = |(1 + \frac{2}{\sqrt{3}}) - (1 - \frac{2}{\sqrt{3}})| = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$.

(C) The equation of the tangent at $R(b, f(b))$ is given by:
$y - f(b) = f'(b)(x - b)$
Since this tangent passes through $S(0, c)$,we have:
$c - f(b) = f'(b)(0 - b)$
$c - f(b) = -b f'(b)$
Given $bc = 3$,we have $c = \frac{3}{b}$. Substituting this into the equation:
$\frac{3}{b} - f(b) = -b f'(b)$
$b f'(b) - f(b) = -\frac{3}{b}$
Dividing both sides by $b^2$:
$\frac{b f'(b) - f(b)}{b^2} = -\frac{3}{b^3}$
This is the derivative of $\frac{f(b)}{b}$ with respect to $b$:
$\frac{d}{db} \left( \frac{f(b)}{b} \right) = -\frac{3}{b^3}$
Integrating both sides with respect to $b$:
$\frac{f(b)}{b} = \int -3b^{-3} db = \frac{3}{2b^2} + \lambda$
$f(b) = \frac{3}{2b} + \lambda b$
Since the curve passes through $P(1, 3/2)$:
$\frac{3}{2} = \frac{3}{2(1)} + \lambda(1) \Rightarrow \lambda = 0$
Thus,$f(x) = \frac{3}{2x}$.
Since the curve passes through $Q(a, 1/2)$:
$f(a) = \frac{3}{2a} = \frac{1}{2} \Rightarrow a = 3$
So,$Q$ is $(3, 1/2)$.
The distance $PQ$ is given by:
$PQ^2 = (3 - 1)^2 + (1/2 - 3/2)^2 = 2^2 + (-1)^2 = 4 + 1 = 5$.

(C) Given the differential equation: $\frac{dT}{dt} = -K(T-80)$.
Separate the variables and integrate: $\int_{160}^{T} \frac{dT}{T-80} = \int_{0}^{t} -K dt$.
This gives: $[ln |T-80|]_{160}^{T} = -Kt$.
$ln |T-80| - ln 80 = -Kt$.
$ln \left| \frac{T-80}{80} \right| = -Kt$,which implies $T-80 = 80e^{-Kt}$,or $T(t) = 80 + 80e^{-Kt}$.
Given $T(15) = 120$,we have $120 = 80 + 80e^{-15K}$.
$40 = 80e^{-15K} \implies e^{-15K} = \frac{40}{80} = \frac{1}{2}$.
We need to find $T(45) = 80 + 80e^{-45K}$.
$T(45) = 80 + 80(e^{-15K})^3$.
Substituting $e^{-15K} = \frac{1}{2}$: $T(45) = 80 + 80 \times (\frac{1}{2})^3$.
$T(45) = 80 + 80 \times \frac{1}{8} = 80 + 10 = 90^{\circ} F$.

(A) Given the equation $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} dt = \int_0^x y(t) dt$.
Applying the Fundamental Theorem of Calculus by differentiating both sides with respect to $x$,we get:
$\sqrt{1-\left(y^{\prime}(x)\right)^2} = y(x)$.
Squaring both sides:
$1-\left(\frac{dy}{dx}\right)^2 = y^2$.
Rearranging the terms:
$\left(\frac{dy}{dx}\right)^2 = 1-y^2$.
Taking the square root:
$\frac{dy}{dx} = \pm \sqrt{1-y^2}$.
Since $y(0)=0$ and $y \geq 0$,we choose the positive root for $x$ near $0$:
$\frac{dy}{\sqrt{1-y^2}} = dx$.
Integrating both sides:
$\sin^{-1}(y) = x + C$.
Using the initial condition $y(0)=0$,we find $C=0$,so $\sin^{-1}(y) = x$,which implies $y = \sin(x)$.
Now,find the derivatives:
$y^{\prime} = \cos(x)$ and $y^{\prime \prime} = -\sin(x)$.
Substituting these into the expression $y^{\prime \prime} + y + 1$:
$-\sin(x) + \sin(x) + 1 = 1$.
Thus,at $x=2$,the value is $1$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$.
Separating the variables,we get: $\int \frac{y}{\sqrt{1-y^2}} dy = \int dx$.
Let $u = 1-y^2$,then $du = -2y dy$,which implies $y dy = -\frac{1}{2} du$.
The integral becomes: $-\frac{1}{2} \int u^{-1/2} du = x + C$.
$-\frac{1}{2} \cdot 2u^{1/2} = x + C \Rightarrow -\sqrt{1-y^2} = x + C$.
Squaring both sides: $1-y^2 = (x+C)^2$.
Rearranging the terms: $(x+C)^2 + y^2 = 1$.
This represents a family of circles with a fixed radius of $1$ and centres at $(-C, 0)$,which lie along the $x$-axis.

(B) The equation of the tangent to the curve $\Gamma$ at point $P(x, y)$ is $Y-y=y^{\prime}(X-x)$.
To find the intersection with the $y$-axis,set $X=0$: $Y_p = y - xy^{\prime}$.
The point $Y_p$ is $(0, y-xy^{\prime})$. The distance $PY_p$ is given as $1$,so the distance between $(x, y)$ and $(0, y-xy^{\prime})$ is $1$.
$\sqrt{(x-0)^2 + (y - (y-xy^{\prime}))^2} = 1$
$\sqrt{x^2 + (xy^{\prime})^2} = 1 \Rightarrow x^2 + x^2(y^{\prime})^2 = 1$
$(y^{\prime})^2 = \frac{1-x^2}{x^2} \Rightarrow y^{\prime} = \pm \frac{\sqrt{1-x^2}}{x}$.
Since the curve is in the first quadrant and passes through $(1,0)$,and the tangent length is constant,the slope must be negative for the curve to decrease towards the $x$-axis. Thus,$y^{\prime} = -\frac{\sqrt{1-x^2}}{x}$.
This gives the differential equation $xy^{\prime} + \sqrt{1-x^2} = 0$,which matches option $(2)$.
Integrating $dy = -\frac{\sqrt{1-x^2}}{x} dx$:
Let $x = \sin\theta$,then $dx = \cos\theta d\theta$.
$y = -\int \frac{\cos\theta}{\sin\theta} \cos\theta d\theta = -\int \frac{\cos^2\theta}{\sin\theta} d\theta = -\int \frac{1-\sin^2\theta}{\sin\theta} d\theta = \int \sin\theta d\theta - \int \csc\theta d\theta$
$y = -\cos\theta - \ln|\csc\theta - \cot\theta| + C = -\sqrt{1-x^2} - \ln\left|\frac{1-\sqrt{1-x^2}}{x}\right| + C$
Using $y(1)=0$,we find $C=0$. Simplifying the logarithm,we get $y = \ln\left(\frac{1+\sqrt{1-x^2}}{x}\right) - \sqrt{1-x^2}$,which matches option $(1)$.

(B) Given $F(x) = \int_0^{x^2} f(\sqrt{t}) dt$. Using the Leibniz rule for differentiation under the integral sign,we get $F'(x) = f(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) = f(x) \cdot 2x$.
Given $F'(x) = f'(x)$,we have $f'(x) = 2x f(x)$.
This is a first-order linear differential equation: $\frac{f'(x)}{f(x)} = 2x$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x^2 + C$.
Since $f(0) = 1$,we have $\ln(1) = 0^2 + C$,which implies $C = 0$.
Thus,$\ln(f(x)) = x^2$,which gives $f(x) = e^{x^2}$.
Now,$F(x) = \int_0^{x^2} e^{(\sqrt{t})^2} dt = \int_0^{x^2} e^t dt = [e^t]_0^{x^2} = e^{x^2} - e^0 = e^{x^2} - 1$.
Therefore,$F(2) = e^{2^2} - 1 = e^4 - 1$.

(A) Let the point $P$ be $(x, y)$. The slope of the tangent at $P$ is $\frac{dy}{dx}$. The slope of the normal at $P$ is $-\frac{dx}{dy}$.
The equation of the normal at $P(x, y)$ is $Y - y = -\frac{dx}{dy}(X - x)$.
To find the point $Q$ where the normal meets the $X$-axis,set $Y = 0$: $-y = -\frac{dx}{dy}(X - x) \implies X - x = y \frac{dy}{dx} \implies X = x + y \frac{dy}{dx}$.
Thus,$Q = (x + y \frac{dy}{dx}, 0)$.
The length $PQ$ is given by $l(PQ)^2 = (X - x)^2 + (0 - y)^2 = k^2$.
Substituting $X - x = y \frac{dy}{dx}$,we get $(y \frac{dy}{dx})^2 + y^2 = k^2$.
$y^2 ((\frac{dy}{dx})^2 + 1) = k^2 \implies (\frac{dy}{dx})^2 = \frac{k^2 - y^2}{y^2} = \frac{k^2 - y^2}{y^2}$.
$\frac{dy}{dx} = \pm \frac{\sqrt{k^2 - y^2}}{y}$.
Separating variables: $\int \frac{y}{\sqrt{k^2 - y^2}} dy = \pm \int dx$.
Let $u = k^2 - y^2$,then $du = -2y dy$. The integral becomes $-\frac{1}{2} \int u^{-1/2} du = -\sqrt{u} = -\sqrt{k^2 - y^2}$.
So,$-\sqrt{k^2 - y^2} = \pm x + C$.
Since the curve passes through $(0, k)$,we have $-\sqrt{k^2 - k^2} = 0 + C \implies C = 0$.
Thus,$\sqrt{k^2 - y^2} = |x| \implies k^2 - y^2 = x^2 \implies x^2 + y^2 = k^2$.

(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.
We know $V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$.
The rate of change of volume is given by $\frac{dV}{dt} = kS$,where $k$ is a constant.
Since $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = (4 \pi r^2) \frac{dr}{dt}$,we have $4 \pi r^2 \frac{dr}{dt} = k(4 \pi r^2)$.
This simplifies to $\frac{dr}{dt} = k$.
Integrating with respect to $t$,we get $r(t) = kt + C$.
At $t = 0$,$r = 3$,so $C = 3$.
At $t = 2$,$r = 9$,so $9 = k(2) + 3$,which gives $2k = 6$,or $k = 3$.
Thus,the radius function is $r(t) = 3t + 3$.
After $t = 4 \ minutes$,$r(4) = 3(4) + 3 = 12 + 3 = 15 \ cm$.

(D) Let $\theta$ be the temperature of the ball at any time $t$. According to Newton's law of cooling,$\frac{d\theta}{dt} = -k(\theta - 20)$,where $k > 0$.
Integrating both sides,we get $\ln|\theta - 20| = -kt + C$.
At $t = 0$,$\theta = 80^{\circ} C$,so $C = \ln(80 - 20) = \ln(60)$.
Thus,$\ln|\theta - 20| = -kt + \ln(60) \dots (i)$.
At $t = 5$,$\theta = 60^{\circ} C$,so $\ln(60 - 20) = -5k + \ln(60)$.
$5k = \ln(60) - \ln(40) = \ln(\frac{60}{40}) = \ln(\frac{3}{2})$.
So,$k = \frac{1}{5} \ln(\frac{3}{2})$.
For $t = 20$,substituting $k$ into equation $(i)$:
$\ln|\theta - 20| = -20 \times \frac{1}{5} \ln(\frac{3}{2}) + \ln(60) = -4 \ln(\frac{3}{2}) + \ln(60)$.
$\ln|\theta - 20| = \ln((\frac{2}{3})^4) + \ln(60) = \ln(\frac{16}{81} \times 60) = \ln(\frac{16 \times 20}{27}) = \ln(\frac{320}{27}) \approx \ln(11.85)$.
$\theta - 20 = 11.85 \implies \theta = 31.85^{\circ} C$.

(C) Let $\theta$ be the temperature of the body at any time $t$. According to Newton's Law of Cooling,the rate of change of temperature is proportional to the difference between the body temperature and the surrounding temperature.
$\frac{d\theta}{dt} = -k(\theta - 30)$
Integrating this,we get $\ln(\theta - 30) = -kt + C$.
At $t = 0$,$\theta = 80^{\circ} C$,so $\ln(80 - 30) = C \Rightarrow C = \ln(50)$.
Thus,$\ln(\theta - 30) = -kt + \ln(50) \dots (i)$.
At $t = 30 \text{ min}$,$\theta = 60^{\circ} C$:
$\ln(60 - 30) = -k(30) + \ln(50) \Rightarrow \ln(30) - \ln(50) = -30k \Rightarrow \ln(3/5) = -30k$.
So,$k = -\frac{1}{30} \ln(3/5) = \frac{1}{30} \ln(5/3)$.
Now,for $t = 60 \text{ min}$ (one hour):
$\ln(\theta - 30) = -\left(\frac{1}{30} \ln(5/3)\right)(60) + \ln(50)$
$\ln(\theta - 30) = -2 \ln(5/3) + \ln(50) = \ln((3/5)^2 \times 50)$
$\ln(\theta - 30) = \ln(9/25 \times 50) = \ln(18)$
$\theta - 30 = 18 \Rightarrow \theta = 48^{\circ} C$.

(A) Let $y$ be the amount of moisture at time $t$.
The rate of change of moisture is proportional to the moisture content:
$\frac{dy}{dt} = -ky$ (where $k > 0$ is a constant).
Separating variables and integrating:
$\int \frac{dy}{y} = -\int k dt \Rightarrow \ln y = -kt + C$.
At $t = 0$,let the initial moisture be $y_0 = 1$ (representing $100 \%$).
Then $\ln(1) = -k(0) + C \Rightarrow C = 0$.
So,$\ln y = -kt$.
Given that at $t = 1$ hour,the sheet loses half its moisture,so $y = 0.5$ (or $1/2$):
$\ln(0.5) = -k(1) \Rightarrow k = -\ln(0.5) = \ln(2)$.
Now,we need to find $t$ when $99 \%$ of the moisture is lost,meaning $1 \%$ remains:
$y = 0.01 = \frac{1}{100} = 10^{-2}$.
Substituting into the equation $\ln y = -kt$:
$\ln(10^{-2}) = -(\ln 2)t$.
$-2 \ln(10) = -(\ln 2)t$.
$t = \frac{2 \ln 10}{\ln 2} = \frac{2 \log 10}{\log 2}$.

(C) Let $p$ be the population at time $t$ years.
Given that the rate of change of population is proportional to the population:
$\frac{dp}{dt} = kp$
$\Rightarrow \frac{dp}{p} = k dt$
Integrating both sides,we get:
$\log p = kt + c$
At $t = 0$,$p = 40,000$:
$\log 40,000 = k(0) + c \Rightarrow c = \log 40,000$
So,$\log p = kt + \log 40,000 \Rightarrow \log \left(\frac{p}{40,000}\right) = kt$
At $t = 40$ years,$p = 80,000$:
$\log \left(\frac{80,000}{40,000}\right) = 40k \Rightarrow \log 2 = 40k \Rightarrow k = \frac{\log 2}{40}$
We need to find the population after another $40$ years,i.e.,at $t = 80$ years:
$\log \left(\frac{p}{40,000}\right) = \left(\frac{\log 2}{40}\right) \times 80$
$\log \left(\frac{p}{40,000}\right) = 2 \log 2 = \log 2^2 = \log 4$
$\frac{p}{40,000} = 4$
$p = 4 \times 40,000 = 160,000$

(B) Let $x$ be the depth of water at time $t$. Given that the rate of flow is proportional to the square root of the depth,we have $\frac{dx}{dt} = -k\sqrt{x}$,where $k > 0$ is a constant (negative sign indicates decreasing depth).
Separating the variables,we get $\frac{dx}{\sqrt{x}} = -k dt$.
Integrating both sides,we get $\int x^{-1/2} dx = \int -k dt$,which gives $2\sqrt{x} = -kt + C$.
At $t = 0$,the initial depth $x = 16 \ m$. Substituting these values,$2\sqrt{16} = -k(0) + C \Rightarrow C = 8$.
So,the equation becomes $2\sqrt{x} = -kt + 8$.
At $t = 2 \ hours$,the depth $x = 4 \ m$. Substituting these values,$2\sqrt{4} = -k(2) + 8 \Rightarrow 4 = -2k + 8 \Rightarrow 2k = 4 \Rightarrow k = 2$.
Thus,the equation is $2\sqrt{x} = -2t + 8$,or $\sqrt{x} = 4 - t$.
For $t = 3.5 \ hours$,$\sqrt{x} = 4 - 3.5 = 0.5$.
Squaring both sides,$x = (0.5)^2 = 0.25 \ m$.

(D) Let $P$ be the population at time $t$. Given $\frac{dP}{dt} \propto P$,so $\frac{dP}{dt} = kP$.
Integrating,we get $\ln P = kt + C$.
At $t = 0$,$P = 20$,so $C = \ln 20$.
Thus,$\ln P = kt + \ln 20$.
At $t = 30$,$P = 40$,so $\ln 40 = 30k + \ln 20$,which gives $30k = \ln 2$,or $k = \frac{\ln 2}{30}$.
We need to find $P$ at $t = 30 + 15 = 45$ years.
$\ln P = \left(\frac{\ln 2}{30}\right) \times 45 + \ln 20 = 1.5 \ln 2 + \ln 20 = \ln(2^{1.5} \times 20)$.
$P = 20 \times 2^{1.5} = 20 \times 2 \times \sqrt{2} = 40 \times 1.41 = 56.4$ lakhs.

(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_s)$,where $T_s = 20^{\circ} C$.
Integrating this gives $T(t) = T_s + (T_0 - T_s)e^{-kt}$.
Given $T_0 = 80^{\circ} C$,we have $T(t) = 20 + 60e^{-kt}$.
For $t = 5$ minutes,$T = 70^{\circ} C$:
$70 = 20 + 60e^{-5k} \Rightarrow 50 = 60e^{-5k} \Rightarrow e^{-5k} = \frac{5}{6}$.
For $t = 15$ minutes:
$T(15) = 20 + 60e^{-15k} = 20 + 60(e^{-5k})^3$.
Substituting $e^{-5k} = \frac{5}{6}$:
$T(15) = 20 + 60 \times \left(\frac{5}{6}\right)^3 = 20 + 60 \times \frac{125}{216} = 20 + \frac{125}{3.6} = 20 + 34.722 = 54.722^{\circ} C$.
Rounding to one decimal place,the temperature is $54.7^{\circ} C$.

(D) Let $m$ be the mass of the radioactive element at time $t$. According to the problem,the rate of disintegration is proportional to its mass:
$\frac{dm}{dt} = -km$ (where $k > 0$ is the decay constant).
Separating the variables,we get:
$\frac{dm}{m} = -k \, dt$.
Integrating both sides:
$\int \frac{dm}{m} = -\int k \, dt \implies \ln(m) = -kt + C$.
At $t = 0$,the initial mass $m = 6 \text{ gm}$,so $\ln(6) = C$.
Thus,$\ln(m) = -kt + \ln(6) \implies \ln(\frac{m}{6}) = -kt$.
We want to find the time $t$ when the mass $m = 3 \text{ gm}$:
$\ln(\frac{3}{6}) = -kt
\implies \ln(\frac{1}{2}) = -kt
\implies -\ln(2) = -kt
\implies t = \frac{\ln(2)}{k}$.
Therefore,the time $t$ is proportional to $\log 2$.

(C) Let $V$ be the volume and $S$ be the surface area of the spherical raindrop. We are given that the rate of evaporation is proportional to the surface area,so $\frac{dV}{dt} = -kS$,where $k > 0$ is a constant.
Since $V = \frac{4}{3}\pi r^3$,we have $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
The surface area $S = 4\pi r^2$.
Substituting these into the rate equation: $4\pi r^2 \frac{dr}{dt} = -k(4\pi r^2)$.
This simplifies to $\frac{dr}{dt} = -k$.
Integrating with respect to $t$,we get $r = -kt + c$.
At $t = 0$,$r = 3$,so $3 = -k(0) + c \Rightarrow c = 3$.
Thus,$r = -kt + 3$.
At $t = 1$,$r = 2$,so $2 = -k(1) + 3 \Rightarrow k = 1$.
Substituting $k=1$ and $c=3$ into the expression for $r$,we get $r = -t + 3$ or $r = 3 - t$.

(A) Let $x$ be the asset at time $t$.
$\frac{dx}{dt} = -k\sqrt{x}$,where $k > 0$.
Separating variables,we get $\frac{dx}{\sqrt{x}} = -k dt$.
Integrating both sides,$2\sqrt{x} = -kt + c$.
At $t = 0$,$x = 10,00,000$,so $2\sqrt{10,00,000} = c \implies c = 2000$.
Thus,$2\sqrt{x} = -kt + 2000$.
At $t = 3$,$x = 10,000$,so $2\sqrt{10,000} = -3k + 2000 \implies 200 = -3k + 2000 \implies 3k = 1800 \implies k = 600$.
For bankruptcy,$x = 0$.
$0 = -600T + 2000 \implies 600T = 2000 \implies T = \frac{2000}{600} = \frac{10}{3}$ years.

(C) Let $x$ be the mass of the material at time $t$.
The rate of decay is given by $\frac{dx}{dt} = -kx$.
Integrating this,we get $\ln(x) = -kt + C$.
At $t=0, x=27$,so $C = \ln(27)$.
Thus,$\ln(x) = -kt + \ln(27)$.
At $t=3, x=8$,so $\ln(8) = -3k + \ln(27)$,which gives $3k = \ln(27) - \ln(8) = \ln(\frac{27}{8})$.
Therefore,$k = \frac{1}{3} \ln(\frac{27}{8}) = \ln((\frac{27}{8})^{1/3}) = \ln(\frac{3}{2})$.
We need to find the mass after one more hour,i.e.,at $t=4$.
$\ln(x) = -4 \ln(\frac{3}{2}) + \ln(27) = \ln((\frac{3}{2})^{-4}) + \ln(27) = \ln(\frac{16}{81} \times 27) = \ln(\frac{16}{3})$.
Therefore,$x = \frac{16}{3} \text{ grams}$.

(C) According to Newton's law of cooling,$\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $\theta_0 = 25^{\circ} C$.
Integrating,we get $\ln(\theta - 25) = -kt + C$.
At $t = 0$,$\theta = 80^{\circ} C$,so $C = \ln(55)$.
Thus,$\ln(\theta - 25) = -kt + \ln(55)$,or $\ln\left(\frac{\theta - 25}{55}\right) = -kt$.
At $t = 30$ min,$\theta = 50^{\circ} C$,so $\ln\left(\frac{50 - 25}{55}\right) = -30k$ $\Rightarrow \ln\left(\frac{25}{55}\right) = -30k$ $\Rightarrow \ln\left(\frac{5}{11}\right) = -30k$.
For $t = 60$ min,$\ln\left(\frac{\theta - 25}{55}\right) = -60k = 2(-30k) = 2 \ln\left(\frac{5}{11}\right)$.
Therefore,$\frac{\theta - 25}{55} = \left(\frac{5}{11}\right)^2 = \frac{25}{121}$.
$\theta - 25 = 55 \times \frac{25}{121} = 5 \times \frac{25}{11} = \frac{125}{11} \approx 11.36$.
$\theta = 25 + 11.36 = 36.36^{\circ} C$.

(A) Let $\theta^{\circ} C$ be the temperature of the body at time $t \text{ min}$. The room temperature is $T_s = 25^{\circ} C$. According to Newton's law of cooling,$\frac{d\theta}{dt} = -k(\theta - T_s)$.
Integrating this,we get $\ln(\theta - 25) = -kt + C$.
At $t = 0$,$\theta = 135^{\circ} C$,so $C = \ln(135 - 25) = \ln(110)$.
Thus,$\ln\left(\frac{\theta - 25}{110}\right) = -kt$.
At $t = 60 \text{ min}$,$\theta = 80^{\circ} C$,so $\ln\left(\frac{80 - 25}{110}\right) = -60k \Rightarrow \ln(0.5) = -60k \Rightarrow k = -\frac{1}{60}\ln(0.5)$.
For $t = 120 \text{ min}$ $(2 \text{ hours})$,$\ln\left(\frac{\theta - 25}{110}\right) = -120 \times \left(-\frac{1}{60}\ln(0.5)\right) = 2\ln(0.5) = \ln(0.5^2) = \ln(0.25)$.
Therefore,$\frac{\theta - 25}{110} = 0.25 \Rightarrow \theta - 25 = 110 \times 0.25 = 27.5$.
$\theta = 27.5 + 25 = 52.5^{\circ} C$.

(C) Given $\frac{d^{2} y}{d x^{2}}=\sin x+e^{x}$.
Integrating both sides with respect to $x$:
$\frac{d y}{d x} = \int (\sin x + e^{x}) dx = -\cos x + e^{x} + c_{1}$.
Given $\frac{d y}{d x} = 4$ at $x=0$:
$4 = -\cos(0) + e^{0} + c_{1} \implies 4 = -1 + 1 + c_{1} \implies c_{1} = 4$.
So,$\frac{d y}{d x} = e^{x} - \cos x + 4$.
Integrating both sides again with respect to $x$:
$y = \int (e^{x} - \cos x + 4) dx = e^{x} - \sin x + 4x + c_{2}$.
Given $y(0) = 3$:
$3 = e^{0} - \sin(0) + 4(0) + c_{2} \implies 3 = 1 - 0 + 0 + c_{2} \implies c_{2} = 2$.
Thus,the equation of the curve is $y = e^{x} - \sin x + 4x + 2$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$
Separating the variables,we get: $\int \frac{y}{\sqrt{1-y^2}} dy = \int dx$
Let $u = 1-y^2$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$.
Substituting this into the integral: $-\frac{1}{2} \int u^{-1/2} du = x + C$
$-\frac{1}{2} \cdot 2u^{1/2} = x + C$
$-\sqrt{1-y^2} = x + C$
Squaring both sides: $1-y^2 = (x+C)^2$
$(x+C)^2 + y^2 = 1$
This is the equation of a circle $(x-h)^2 + (y-k)^2 = r^2$,where the centre is $(-C, 0)$ and the radius $r = 1$.
Since $C$ is an arbitrary constant,the centre $(-C, 0)$ varies along the $X$-axis,while the radius remains fixed at $1$ unit.

(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_s)$,where $T_s = 25^{\circ} C$.
Integrating this,we get $T(t) = T_s + (T_0 - T_s)e^{-kt}$.
Given $T_0 = 100^{\circ} C$,so $T(t) = 25 + 75e^{-kt}$.
At $t = 10 \text{ min}$,$T = 80^{\circ} C$:
$80 = 25 + 75e^{-10k} \Rightarrow 55 = 75e^{-10k} \Rightarrow e^{-10k} = \frac{55}{75} = \frac{11}{15}$.
At $t = 20 \text{ min}$,$T = 25 + 75e^{-20k} = 25 + 75(e^{-10k})^2$.
Substituting the value: $T = 25 + 75 \times (\frac{11}{15})^2 = 25 + 75 \times \frac{121}{225} = 25 + \frac{121}{3} = 25 + 40.33 = 65.33^{\circ} C$.

(B) Given the differential equation: $x \frac{d^2 y}{d x^2} = 1$.
Dividing by $x$ (assuming $x \neq 0$): $\frac{d^2 y}{d x^2} = \frac{1}{x}$.
Integrating with respect to $x$: $\frac{dy}{dx} = \int \frac{1}{x} dx = \log x + C_1$.
Given $\frac{dy}{dx} = 0$ at $x = 1$: $0 = \log(1) + C_1 \implies C_1 = 0$.
So,$\frac{dy}{dx} = \log x$.
Integrating again with respect to $x$: $y = \int \log x dx = x \log x - x + C_2$.
Given $y = 1$ at $x = 1$: $1 = 1 \log(1) - 1 + C_2 \implies 1 = 0 - 1 + C_2 \implies C_2 = 2$.
Thus,the solution is $y = x \log x - x + 2$.

(B) Given the slope of the curve is $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.
Integrating both sides with respect to $x$:
$\int dy = \int \left(1 - \frac{1}{x^2}\right) dx$
$y = x + \frac{1}{x} + C$
Since the curve passes through $\left(2, \frac{9}{2}\right)$,substitute $x = 2$ and $y = \frac{9}{2}$:
$\frac{9}{2} = 2 + \frac{1}{2} + C$
$\frac{9}{2} = \frac{5}{2} + C$
$C = \frac{9}{2} - \frac{5}{2} = 2$
Thus,the equation is $y = x + \frac{1}{x} + 2$.
Multiplying by $x$,we get $xy = x^2 + 1 + 2x$,which simplifies to $xy = x^2 + 2x + 1$.

(C) Let $M(t)$ be the moisture content at time $t$. The rate of change is given by $\frac{dM}{dt} = -kM$,where $k > 0$.
Solving this differential equation,we get $M(t) = M_0 e^{-kt}$,where $M_0$ is the initial moisture.
Given that at $t = 1$ hour,the moisture lost is $50 \%$,so $M(1) = 0.5 M_0$.
$0.5 M_0 = M_0 e^{-k} \implies e^{-k} = 0.5 = \frac{1}{2}$.
Taking natural logarithms,$-k = \ln(1/2) = -\ln 2$,so $k = \ln 2$.
We want to find $t$ such that $90 \%$ of the moisture is lost,meaning $10 \%$ remains.
$M(t) = 0.1 M_0 = M_0 e^{-kt}$.
$0.1 = e^{-kt} \implies \ln(0.1) = -kt$.
$-\ln 10 = -(\ln 2)t$.
$t = \frac{\ln 10}{\ln 2} = \log_2 10$ hours.

(B) Given the differential equation: $x \frac{dy}{dx} = 2y$.
Separating the variables,we get: $\frac{dy}{y} = 2 \frac{dx}{x}$.
Integrating both sides: $\int \frac{dy}{y} = 2 \int \frac{dx}{x}$.
This gives: $\ln|y| = 2 \ln|x| + C$.
Using properties of logarithms: $\ln|y| = \ln|x^2| + C$.
Taking the exponential of both sides: $y = e^{\ln|x^2| + C} = e^C \cdot x^2$.
Let $e^C = k$,where $k$ is an arbitrary constant.
Thus,$y = kx^2$.
This equation represents a family of parabolas with the vertex at the origin $(0,0)$ and the axis along the $Y$-axis.

(B) Given the differential equation: $\frac{dp}{dt} = \frac{p - 200}{2}$.
Separating the variables,we get: $\frac{dp}{p - 200} = \frac{1}{2} dt$.
Integrating both sides: $\int \frac{dp}{p - 200} = \int \frac{1}{2} dt$.
This gives: $\ln|p - 200| = \frac{t}{2} + C$.
Exponentiating both sides: $p - 200 = e^{t/2 + C} = Ae^{t/2}$,where $A = \pm e^C$.
So,$p = 200 + Ae^{t/2}$.
Given the initial condition $p(0) = 100$:
$100 = 200 + Ae^0 \implies 100 = 200 + A \implies A = -100$.
Substituting $A$ back into the equation: $p = 200 - 100 e^{t/2}$.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$
Separate the variables: $\int \frac{y}{\sqrt{1-y^2}} dy = \int dx$
Let $u = 1-y^2$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$.
The integral becomes: $-\frac{1}{2} \int u^{-1/2} du = x + C$
$-\frac{1}{2} (2u^{1/2}) = x + C$
$-\sqrt{1-y^2} = x + C$
Square both sides: $1-y^2 = (x+C)^2$
$(x+C)^2 + y^2 = 1$
This is the equation of a circle $(x-h)^2 + (y-k)^2 = r^2$ where $h = -C$,$k = 0$,and $r = 1$.
Thus,the radius is fixed at $1$ unit and the centre $(-C, 0)$ varies along the $X$-axis.

(B) According to Newton's Law of Cooling,the rate of change of temperature of a body is proportional to the difference between the temperature of the body $T$ and the temperature of the surrounding medium $T_s$.
Here,$T_s = 32^{\circ} F$.
Since the body is hot,its temperature $T$ decreases as time $t$ increases,so $\frac{dT}{dt} < 0$.
Thus,$\frac{dT}{dt} \propto -(T - 32)$.
Introducing a positive constant of proportionality $k$,we get:
$\frac{dT}{dt} = -k(T - 32)$.

(A) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.
The slope of the normal is $-\frac{dx}{dy}$.
According to the problem,the slope of the normal is equal to the ordinate $y$,so $-\frac{dx}{dy} = y$.
This implies $dx = -y \, dy$.
Integrating both sides,we get $x = -\frac{y^2}{2} + C$.
Since the curve passes through $(1, -1)$,we substitute these values: $1 = -\frac{(-1)^2}{2} + C$,which gives $1 = -\frac{1}{2} + C$,so $C = \frac{3}{2}$.
Thus,the equation is $x = -\frac{y^2}{2} + \frac{3}{2}$,which simplifies to $2x = -y^2 + 3$,or $2x = 1(3 - y^2)$.
Comparing this with $2x = k(3 - y^2)$,we find $k = 1$.

(A) Let $P$ be the principal at time $t$. Given that the principal increases continuously at a rate of $x \%$ per year,we have the differential equation: $\frac{dP}{dt} = \frac{x}{100} P$.
Integrating this,we get $\int \frac{dP}{P} = \int \frac{x}{100} dt$,which gives $\ln P = \frac{x}{100} t + C$.
At $t = 0$,$P = P_0 = 100$. Thus,$C = \ln 100$.
So,$\ln P = \frac{x}{100} t + \ln 100$,or $\ln(\frac{P}{100}) = \frac{x}{100} t$.
Given that the principal doubles in $10$ years,at $t = 10$,$P = 200$.
Substituting these values: $\ln(\frac{200}{100}) = \frac{x}{100} \times 10$.
$\ln 2 = \frac{x}{10}$.
Since $\ln 2 = 2.3026 \times \log_{10} 2$,we have $\ln 2 = 2.3026 \times 0.6931 \approx 1.596$.
However,if the rate is defined such that $\frac{dP}{dt} = rP$ where $r = \frac{x}{100}$,then $P(t) = P_0 e^{rt}$.
$200 = 100 e^{10r} \implies 2 = e^{10r} \implies 10r = \ln 2$.
Using $\ln 2 \approx 0.6931$,$10r = 0.6931 \implies r = 0.06931$.
Thus,$x = 100r = 6.931 \% \approx 6.93 \%$.

(B) Let $P$ be the principal at any time $t$.
Given that the principal increases continuously at the rate of $10 \%$ per year,we have the differential equation:
$\frac{dP}{dt} = 0.10 P$
Separating the variables,we get:
$\frac{dP}{P} = 0.10 dt$
Integrating both sides:
$\int \frac{dP}{P} = \int 0.10 dt$
$\ln(P) = 0.10 t + C$
$P(t) = e^{0.10 t + C} = Ae^{0.10 t}$
At $t = 0$,$P = 2000$. Substituting these values:
$2000 = Ae^{0.10(0)} \implies A = 2000$
So,the equation for the principal is $P(t) = 2000 e^{0.10 t}$.
After $t = 5$ years:
$P(5) = 2000 e^{0.10(5)} = 2000 e^{0.5}$
Given $e^{0.5} = 1.648$:
$P(5) = 2000 \times 1.648 = 3296$
Thus,the amount after $5$ years will be Rs. $3296$.

(C) Let $A(t)$ be the assets at time $t$. The rate of reduction is proportional to the square root of the assets, so $\frac{dA}{dt} = -k \sqrt{A}$, where $k > 0$.
Separating variables, we get $\frac{dA}{\sqrt{A}} = -k dt$.
Integrating both sides, $\int A^{-1/2} dA = \int -k dt$, which gives $2\sqrt{A} = -kt + C$.
At $t = 0$, $A = 25$, so $2\sqrt{25} = C \implies C = 10$.
Thus, $2\sqrt{A} = -kt + 10$.
At $t = 2$, $A = 6.25$, so $2\sqrt{6.25} = -k(2) + 10$.
$2(2.5) = -2k + 10 \implies 5 = -2k + 10 \implies 2k = 5 \implies k = 2.5$.
The equation becomes $2\sqrt{A} = -2.5t + 10$.
For bankruptcy, $A = 0$, so $0 = -2.5t + 10$.
$2.5t = 10 \implies t = \frac{10}{2.5} = 4 \text{ years}$.

(B) Let $A(t)$ be the assets at time $t$. The rate of reduction is proportional to the square root of the assets,so $\frac{dA}{dt} = -k\sqrt{A}$,where $k > 0$.
Separating variables,we get $A^{-1/2} dA = -k dt$.
Integrating both sides,we get $2\sqrt{A} = -kt + C$.
At $t = 0$,$A = 10,00,000$. Thus,$2\sqrt{10,00,000} = C \implies C = 2 \times 1000 = 2000$.
So,$2\sqrt{A} = -kt + 2000$.
At $t = 3$,$A = 10,000$. Thus,$2\sqrt{10,000} = -3k + 2000 \implies 200 = -3k + 2000 \implies 3k = 1800 \implies k = 600$.
The equation is $2\sqrt{A} = -600t + 2000$.
For bankruptcy,$A = 0$,so $0 = -600t + 2000$.
$600t = 2000 \implies t = \frac{2000}{600} = \frac{20}{6} = \frac{10}{3}$ years.

(A) Let $N$ be the number of bacteria at time $t$. Given that the rate of growth is proportional to the number present,we have $\frac{dN}{dt} = kN$.
Integrating this,we get $\ln N = kt + C$,or $N(t) = N_0 e^{kt}$.
Initially,at $t = 0$,$N_0 = 1,00,000$.
After $2$ hours,the count increases by $10 \%$,so $N(2) = 1,00,000 + 0.10 \times 1,00,000 = 1,10,000$.
Substituting these values: $1,10,000 = 1,00,000 e^{2k} \implies e^{2k} = 1.1 \implies 2k = \ln(1.1) \implies k = \frac{\ln(1.1)}{2}$.
We want to find $t$ such that $N(t) = 2,00,000$.
$2,00,000 = 1,00,000 e^{kt} \implies 2 = e^{kt} \implies \ln 2 = kt$.
Substituting $k$: $\ln 2 = \left(\frac{\ln(1.1)}{2}\right) t$.
Solving for $t$: $t = \frac{2 \ln 2}{\ln(1.1)} = \frac{2 \log 2}{\log(1.1)}$.

(C) Let $P(t)$ be the population at time $t$. The rate of change is given by $\frac{dP}{dt} = kP$.
Integrating this,we get $\ln P = kt + C$,or $P(t) = P_0 e^{kt}$.
At $t = 0$,$P(0) = 40,000$. So,$P_0 = 40,000$.
At $t = 20$,$P(20) = 80,000$. Thus,$80,000 = 40,000 e^{20k}$,which implies $e^{20k} = 2$.
We want to find the population after another $40$ years,which is at $t = 20 + 40 = 60$ years.
$P(60) = 40,000 e^{60k} = 40,000 (e^{20k})^3$.
Substituting $e^{20k} = 2$,we get $P(60) = 40,000 \times (2)^3 = 40,000 \times 8 = 320,000$.

(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = -k(T - T_a)$,where $T_a = 290 \ K$.
Integrating,we get $\ln(T - 290) = -kt + C$,or $T - 290 = Ce^{-kt}$.
At $t = 0$,$T = 370$,so $370 - 290 = C$,which gives $C = 80$.
Thus,$T - 290 = 80e^{-kt}$.
At $t = 10 \ min$,$T = 330$,so $330 - 290 = 80e^{-10k}$,which means $40 = 80e^{-10k}$,so $e^{-10k} = 0.5$.
Taking natural logs,$-10k = \ln(0.5) = -\ln(2)$,so $k = \frac{\ln(2)}{10}$.
We want to find $t$ when $T = 295$.
$295 - 290 = 80e^{-kt} \implies 5 = 80e^{-kt} \implies e^{-kt} = \frac{5}{80} = \frac{1}{16} = (\frac{1}{2})^4$.
Since $e^{-10k} = \frac{1}{2}$,we have $e^{-kt} = (e^{-10k})^4 = e^{-40k}$.
Therefore,$kt = 40k$,which gives $t = 40 \ minutes$.

(C) Let $P(t)$ be the population at time $t$. According to the problem,the rate of increase is proportional to the population: $\frac{dP}{dt} = kP$.
Integrating this,we get $\ln P = kt + C$,or $P(t) = P_0 e^{kt}$.
At $t = 0$,$P(0) = 4$ lakhs,so $P_0 = 4$.
At $t = 20$,$P(20) = 6$ lakhs. Thus,$6 = 4 e^{20k}$,which gives $e^{20k} = \frac{6}{4} = 1.5$.
We need to find the population after another $20$ years,i.e.,at $t = 40$.
$P(40) = P_0 e^{40k} = 4 (e^{20k})^2$.
Substituting $e^{20k} = 1.5$,we get $P(40) = 4 \times (1.5)^2 = 4 \times 2.25 = 9$ lakhs.

(B) Let $P(t)$ be the population at time $t$. According to the problem,$\frac{dP}{dt} \propto P$,which implies $\frac{dP}{dt} = kP$.
Solving this differential equation,we get $P(t) = Ce^{kt}$.
At $t = 0$,the population is $30,000$,so $P(0) = C = 30,000$.
Thus,$P(t) = 30,000 e^{kt}$.
At $t = 40$,the population is $40,000$,so $40,000 = 30,000 e^{40k}$.
This simplifies to $e^{40k} = \frac{40,000}{30,000} = \frac{4}{3}$.
Substituting this into the equation for $P(t)$,we get $P(t) = 30,000 (e^{40k})^{\frac{t}{40}} = 30,000 (\frac{4}{3})^{\frac{t}{40}}$.
Comparing this with the given form $P(t) = (a)(b)^{\frac{t}{40}}$,we find $a = 30,000$ and $b = \frac{4}{3}$.

(B) For continuous compounding,the amount $A$ at time $t$ is given by the formula $A(t) = P e^{rt}$,where $P$ is the principal amount,$r$ is the rate of interest,and $t$ is the time in years.
Given $P = 400$ and $A(6) = 800$,we have $800 = 400 e^{6r}$.
Dividing by $400$,we get $2 = e^{6r}$.
Taking the natural logarithm on both sides,$\ln(2) = 6r$,so $r = \frac{\ln(2)}{6}$.
We need to find the amount at $t = 30$ years: $A(30) = 400 e^{30r}$.
Substituting $r = \frac{\ln(2)}{6}$,we get $A(30) = 400 e^{30 \times \frac{\ln(2)}{6}} = 400 e^{5 \ln(2)}$.
Using the property $e^{a \ln(b)} = b^a$,we get $A(30) = 400 \times 2^5$.
$A(30) = 400 \times 32 = 12800$.
Thus,the amount at the end of $30$ years will be ₹ $12800$.