Mix Examples-Differential Equations Questions in English

(B) Given differential equation: $(1+xy)y \, dx + (1-xy)x \, dy = 0$
Expanding the terms: $y \, dx + xy^2 \, dx + x \, dy - x^2y \, dy = 0$
Rearranging: $(y \, dx + x \, dy) + xy(y \, dx - x \, dy) = 0$
Dividing by $x^2y^2$: $\frac{y \, dx + x \, dy}{x^2y^2} + \frac{y \, dx - x \, dy}{xy} = 0$
This can be written as: $\frac{d(xy)}{(xy)^2} - \left(\frac{x \, dy - y \, dx}{xy}\right) = 0$
Integrating both sides: $\int \frac{d(xy)}{(xy)^2} - \int d\left(\log \frac{x}{y}\right) = \int 0$
$-\frac{1}{xy} - \log \left(\frac{x}{y}\right) = -k$
$\log \left(\frac{x}{y}\right) = \frac{1}{xy} + k$

(A) Given the determinant equation: $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$
Expanding the determinant,we get: $f(x) f^{\prime \prime}(x) - (f^{\prime}(x))^2 = 0$
$\Rightarrow f(x) f^{\prime \prime}(x) = (f^{\prime}(x))^2$
Dividing both sides by $f(x) f^{\prime}(x)$ (assuming $f(x), f^{\prime}(x) \neq 0$): $\frac{f^{\prime \prime}(x)}{f^{\prime}(x)} = \frac{f^{\prime}(x)}{f(x)}$
Integrating both sides with respect to $x$: $\int \frac{f^{\prime \prime}(x)}{f^{\prime}(x)} dx = \int \frac{f^{\prime}(x)}{f(x)} dx$
$\ln(f^{\prime}(x)) = \ln(f(x)) + C_1$
Using initial conditions $f(0)=1$ and $f^{\prime}(0)=2$: $\ln(2) = \ln(1) + C_1 \Rightarrow C_1 = \ln(2)$
So,$\ln(f^{\prime}(x)) = \ln(f(x)) + \ln(2) = \ln(2f(x))$
$\Rightarrow f^{\prime}(x) = 2f(x)$
$\Rightarrow \frac{f^{\prime}(x)}{f(x)} = 2$
Integrating again: $\int \frac{f^{\prime}(x)}{f(x)} dx = \int 2 dx$
$\ln(f(x)) = 2x + C_2$
Using $f(0)=1$: $\ln(1) = 2(0) + C_2 \Rightarrow C_2 = 0$
Thus,$\ln(f(x)) = 2x \Rightarrow f(x) = e^{2x}$
Therefore,the correct option is $A$.

(A) Given the differential equation: $\frac{d^2 y}{d x^2}+y=0$.
We check if $y=3 \sin x+4 \cos x$ is a solution.
Differentiating with respect to $x$:
$\frac{d y}{d x}=3 \cos x-4 \sin x$.
Differentiating again with respect to $x$:
$\frac{d^2 y}{d x^2}=-3 \sin x-4 \cos x$.
Substituting this into the original equation:
$\frac{d^2 y}{d x^2}+y = (-3 \sin x-4 \cos x) + (3 \sin x+4 \cos x) = 0$.
Since the equation is satisfied,$y=3 \sin x+4 \cos x$ is a solution.
Alternative Method:
The characteristic equation for $\frac{d^2 y}{d x^2}+y=0$ is $m^2+1=0$,which gives $m = \pm i$.
The general solution is $y=c_1 \cos x+c_2 \sin x$.
Option $A$ is of this form with $c_1=4$ and $c_2=3$.

(C) Let the slope of the first curve be $m_1 = \frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$.
Let the slope of the second curve be $m_2 = \frac{dy}{dx} = -\frac{y^3 e^{y^2-5}}{x \log x}$.
Now,calculate the product of the slopes: $m_1 \times m_2 = \left(\frac{x \log x}{y^3 e^{y^2-5}}\right) \times \left(-\frac{y^3 e^{y^2-5}}{x \log x}\right) = -1$.
Since the product of the slopes is $-1$,the curves are orthogonal,meaning the angle of intersection is $\theta = \frac{\pi}{2}$.
Finally,calculate $4\theta - \frac{\pi}{2} = 4\left(\frac{\pi}{2}\right) - \frac{\pi}{2} = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.

(D) Given the differential equation: $y dx - x dy + 3x^2 y^2 e^{x^3} dx = 0$.
Dividing the entire equation by $y^2$,we get: $\frac{y dx - x dy}{y^2} + 3x^2 e^{x^3} dx = 0$.
This can be written as: $d(\frac{x}{y}) + d(e^{x^3}) = 0$.
Integrating both sides,we obtain: $\frac{x}{y} + e^{x^3} = C$,where $C$ is the constant of integration.
Given that $y = 1$ when $x = 1$,we substitute these values: $\frac{1}{1} + e^{1^3} = C \Rightarrow C = 1 + e$.
Substituting the value of $C$ back into the equation: $\frac{x}{y} + e^{x^3} = 1 + e$.
Rearranging the terms: $\frac{x}{y} = 1 + e - e^{x^3} \Rightarrow x = y(1 + e - e^{x^3}) \Rightarrow x = y(1 + e - e^{x^3}) \Rightarrow x = -y(e^{x^3} - (1 + e)) \Rightarrow x + y(e^{x^3} - (1 + e)) = 0$.

(D) Given differential equation: $x y \, dy - (1 + y^2) \, dx = 0$.
Separating the variables,we get $\frac{y}{1 + y^2} \, dy = \frac{1}{x} \, dx$.
Integrating both sides: $\int \frac{y}{1 + y^2} \, dy = \int \frac{1}{x} \, dx$.
$\frac{1}{2} \ln(1 + y^2) = \ln|x| + C$.
Since the curve passes through $(1, 0)$,we have $\frac{1}{2} \ln(1 + 0) = \ln(1) + C \Rightarrow C = 0$.
Thus,$\ln(1 + y^2) = 2 \ln|x| \Rightarrow 1 + y^2 = x^2$.
For the curve $x^2 - y^2 = 1$,differentiating w.r.t. $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow m_1 = \frac{dy}{dx} = \frac{x}{y}$.
For the curve $x^2 + 3y^2 = 3$,differentiating w.r.t. $x$: $2x + 6y \frac{dy}{dx} = 0 \Rightarrow m_2 = \frac{dy}{dx} = -\frac{x}{3y}$.
At the intersection point,$x^2 = 1 + y^2$. Substituting into $x^2 + 3y^2 = 3$: $(1 + y^2) + 3y^2 = 3 \Rightarrow 4y^2 = 2 \Rightarrow y^2 = \frac{1}{2}$.
Then $x^2 = 1 + \frac{1}{2} = \frac{3}{2}$.
At the intersection point,$m_1 m_2 = (\frac{x}{y})(-\frac{x}{3y}) = -\frac{x^2}{3y^2} = -\frac{3/2}{3(1/2)} = -1$.
Since the product of slopes is $-1$,the curves intersect at an angle $\theta = \frac{\pi}{2}$.
Therefore,$\frac{2\theta}{\pi} = \frac{2(\pi/2)}{\pi} = 1$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{x+y+1}{y-x+1}$.
Cross-multiplying,we get $(y-x+1) dy = (x+y+1) dx$.
Rearranging the terms: $y dy - x dy + dy = x dx + y dx + dx$.
Grouping terms: $(y+1) dy - x dy = (x+1) dx + y dx$.
Adding $x dy$ to both sides: $(y+1) dy = (x+1) dx + (y dx + x dy)$.
Recognizing the product rule $d(xy) = y dx + x dy$,we have $(y+1) dy = (x+1) dx + d(xy)$.
Integrating both sides: $\int (y+1) dy = \int (x+1) dx + \int d(xy)$.
This yields $\frac{(y+1)^2}{2} = \frac{(x+1)^2}{2} + xy + C_1$.
Multiplying by $2$: $(y+1)^2 = (x+1)^2 + 2xy + 2C_1$.
Rearranging: $2xy + (x+1)^2 - (y+1)^2 = -2C_1$.
Letting $C = -2C_1$,we get $2xy + (x+1)^2 - (y+1)^2 = C$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{2x-3y+4}{3x+2y-7}$
Cross-multiplying,we get: $(3x+2y-7)dy = (2x-3y+4)dx$
Rearranging the terms: $(3x+2y-7)dy - (2x-3y+4)dx = 0$
$(3xdy + 3ydx) + (2ydy - 2xdx) - 7dy - 4dx = 0$
This can be written as: $3d(xy) + d(y^2) - d(x^2) - 7dy - 4dx = 0$
Integrating both sides: $\int 3d(xy) + \int d(y^2) - \int d(x^2) - \int 7dy - \int 4dx = \int 0$
$3xy + y^2 - x^2 - 7y - 4x = C$
Multiplying by $-1$: $x^2 - y^2 - 3xy + 4x + 7y + C = 0$

(C) Given the differential equation: $(y-x+1) dy - (y+x+2) dx = 0$.
Rearranging the terms,we have: $y dy - x dy + dy - y dx - x dx + 2 dx = 0$.
Grouping the terms: $y dy + dy - (x dy + y dx) - x dx + 2 dx = 0$.
Recognizing the exact differential $d(xy) = x dy + y dx$,the equation becomes: $y dy + dy - d(xy) - x dx + 2 dx = 0$.
Integrating both sides: $\int y dy + \int dy - \int d(xy) - \int x dx + \int 2 dx = \int 0$.
This yields: $\frac{y^2}{2} + y - xy - \frac{x^2}{2} + 2x = C$.
Thus,$f(x, y, c) = \frac{y^2}{2} - \frac{x^2}{2} - xy + 2x + y - C = 0$.
Given $f(1, 1, c) = 0$,substitute $x = 1$ and $y = 1$:
$\frac{1^2}{2} - \frac{1^2}{2} - (1)(1) + 2(1) + 1 - C = 0$.
$\frac{1}{2} - \frac{1}{2} - 1 + 2 + 1 - C = 0$.
$2 - C = 0$,which implies $C = 2$.

(B) Given the transformation $z = \log \tan \frac{x}{2}$.
First,find $\frac{d z}{d x}$:
$\frac{d z}{d x} = \frac{1}{\tan \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \operatorname{cosec} x$.
Now,express $\frac{d y}{d z}$ in terms of $x$:
$\frac{d y}{d z} = \frac{d y}{d x} \cdot \frac{d x}{d z} = \frac{d y}{d x} \cdot \sin x$.
Next,find $\frac{d^2 y}{d z^2}$:
$\frac{d^2 y}{d z^2} = \frac{d}{d z} \left( \sin x \frac{d y}{d x} \right) = \frac{d}{d x} \left( \sin x \frac{d y}{d x} \right) \cdot \frac{d x}{d z} = \left( \cos x \frac{d y}{d x} + \sin x \frac{d^2 y}{d x^2} \right) \cdot \sin x = \sin x \cos x \frac{d y}{d x} + \sin^2 x \frac{d^2 y}{d x^2}$.
Substitute this into the target equation $\frac{d^2 y}{d z^2} + k y = 0$:
$\sin^2 x \frac{d^2 y}{d x^2} + \sin x \cos x \frac{d y}{d x} + k y = 0$.
Divide by $\sin^2 x$:
$\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + k \operatorname{cosec}^2 x \cdot y = 0$.
Comparing this with the given equation $\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + 4 \operatorname{cosec}^2 x \cdot y = 0$,we get $k = 4$.

(B) Given $y = \frac{x}{\ln|cx|}$.
Taking the derivative with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{\ln|cx| \cdot 1 - x \cdot \frac{1}{cx} \cdot c}{(\ln|cx|)^2} = \frac{\ln|cx| - 1}{(\ln|cx|)^2} = \frac{1}{\ln|cx|} - \frac{1}{(\ln|cx|)^2}$.
Since $y = \frac{x}{\ln|cx|}$,we have $\frac{y}{x} = \frac{1}{\ln|cx|}$.
Substituting this into the derivative expression:
$\frac{dy}{dx} = \frac{y}{x} - \left(\frac{y}{x}\right)^2$.
Comparing this with the given differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi\left(\frac{x}{y}\right)$,we identify $\phi\left(\frac{x}{y}\right) = -\left(\frac{y}{x}\right)^2 = -\frac{y^2}{x^2}$.

(B) Given the differential equation: $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$.
Rearranging the terms,we get: $\frac{1 - y \log x}{y^2} dy = \frac{1}{x} dx$.
Integrating both sides: $\int (y^{-2} - \frac{\log x}{y}) dy = \int \frac{1}{x} dx$ is not direct. Let us check the options.
If $y = x^y$,then taking $\log$ on both sides: $\log y = y \log x$.
Differentiating w.r.t. $x$: $\frac{1}{y} \frac{dy}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}$.
Rearranging for $\frac{dy}{dx}$: $\frac{dy}{dx} (\frac{1}{y} - \log x) = \frac{y}{x}$.
$\frac{dy}{dx} (\frac{1 - y \log x}{y}) = \frac{y}{x}$.
$x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$.
This matches the given differential equation. Also,for $x=1$,$y=1^y=1$,which satisfies the condition $y(1)=1$.

(B) Given $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$.
Using the Leibniz integral rule,we differentiate with respect to $x$:
$f'(x) = \tan(x-x) - \tan(0-x) - f(x) \tan x = 0 - (-\tan x) - f(x) \tan x = \tan x (1 - f(x))$.
This is a separable differential equation: $\frac{df}{1-f} = \tan x dx$.
Integrating both sides: $-\ln|1-f| = \ln|\sec x| + C$.
This simplifies to $\ln|1-f|^{-1} = \ln|\sec x| + C$,or $1-f = k \cos x$.
At $x=0$,$f(0) = \int_{0}^{0} \tan(t) dt - \int_{0}^{0} f(t) \tan t dt = 0$.
Substituting $x=0$ into $1-f(x) = k \cos x$,we get $1-0 = k(1)$,so $k=1$.
Thus,$f(x) = 1 - \cos x$.
Now,$f'(x) = \sin x$ and $f''(x) = \cos x$.
We need to evaluate $f''(\pi/6) + f(\pi/6)$:
$f''(\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}$.
$f(\pi/6) = 1 - \cos(\pi/6) = 1 - \frac{\sqrt{3}}{2}$.
Therefore,$f''(\pi/6) + f(\pi/6) = \frac{\sqrt{3}}{2} + 1 - \frac{\sqrt{3}}{2} = 1$.