Differential Equations
A function $y=f(x)$ satisfies $f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0$ with condition $f(0)=0$. Then, $f\left(\frac{\pi}{2}\right)$ is equal to
If $y=y(x)$ is the solution curve of the differential equation $\left(x^2-4\right) \mathrm{d} y-\left(y^2-3 y\right) \mathrm{d} x=0, x>2, y(4)=\frac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals :
For a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$, suppose $f^{\prime}(x)=3 f(x)+\alpha$, where $\alpha \in \mathbb{R}, f(0)=1$ and $\lim _\limits{x \rightarrow-\infty} f(x)=7$. Then $9 f\left(-\log _e 3\right)$ is equal to _________.
Explanation:
$\begin{aligned} & f^{\prime}(x)=3 f(x)+\alpha \\ & \Rightarrow \frac{d y}{3 y+\alpha}=d x \\ & \Rightarrow \frac{1}{3} \ln (3 y+\alpha)=x+C \\ & y(0)=1 \Rightarrow C=\frac{1}{3} \ln (3+\alpha) \\ & \frac{1}{3} \ln \left(\frac{3 y+\alpha}{3+\alpha}\right)=x \\ & \Rightarrow y=\frac{1}{3}\left((3+\alpha) e^{3 x}-\alpha\right)=f(x) \\ & \lim _{x \rightarrow-\infty} f(x)=7 \Rightarrow \alpha=-21 \\ & \Rightarrow f(x)=7-6 e^{3 x} \\ & 9 f(-\ln 3)=61 \end{aligned}$
Let $\alpha|x|=|y| \mathrm{e}^{x y-\beta}, \alpha, \beta \in \mathbf{N}$ be the solution of the differential equation $x \mathrm{~d} y-y \mathrm{~d} x+x y(x \mathrm{~d} y+y \mathrm{~d} x)=0,y(1)=2$. Then $\alpha+\beta$ is equal to ________
Explanation:
$\begin{aligned} & \alpha|x|=|y| e^{x y-\beta} \\ & \frac{x d y-y d x}{y^2}+\frac{x y(x d y+y d x)}{y^2}=0 \\ & -d\left(\frac{x}{y}\right)+\frac{x}{y} d(x y)=0 \end{aligned}$
$\begin{aligned} & \int d(x y)=\int \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}} \\ & x y=\ln \left|\frac{x}{y}\right|+\ln c \\ & x y=\ln \left(\left|\frac{x}{y}\right| \cdot c\right) \\ & \because y(1)=2 \\ & 2=\ln \left|\frac{1}{2}\right| c \Rightarrow c=2 e^2 \\ & \therefore \quad \operatorname{solution} x y=\ln \left(\left|\frac{x}{y}\right| \cdot 2 e^2\right) \\ & e^{x y}=\frac{|x|}{|y|} \cdot 2 e^2 \\ & 2|x|=|y| e^{x y-2} \\ & \Rightarrow \alpha=2, \beta=2, \alpha+\beta=4 \end{aligned}$
If the solution $y(x)$ of the given differential equation $\left(e^y+1\right) \cos x \mathrm{~d} x+\mathrm{e}^y \sin x \mathrm{~d} y=0$ passes through the point $\left(\frac{\pi}{2}, 0\right)$, then the value of $e^{y\left(\frac{\pi}{6}\right)}$ is equal to _________.
Explanation:
Given the differential equation
$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$
we aim to find the value of $ e^{y\left(\frac{\pi}{6}\right)} $ given that the solution $ y(x) $ passes through the point $\left(\frac{\pi}{2}, 0\right)$.
First, we recognize that the differential equation can be rearranged as:
$d\left(e^y \sin x\right) + \cos x \, dx = 0.$
Integrating this expression, we obtain:
$ e^y \sin x + \sin x = C,$
where $ C $ is a constant. Given that the solution passes through the point $\left(\frac{\pi}{2}, 0\right)$, we substitute these values into the equation to find $ C $:
$ e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C \Rightarrow 1 + 1 = C \Rightarrow C = 2.$
Thus, the equation simplifies to:
$e^y \sin x + \sin x = 2.$
We now need to determine the value of $ e^{y\left(\frac{\pi}{6}\right)} $. Substituting $ x = \frac{\pi}{6} $ into the equation, we get:
$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \sin\left(\frac{\pi}{6}\right) = 2.$
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the equation becomes:
$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \frac{1}{2} = 2 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} + 1 = 4 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} = 3.$
Therefore, the value of $ e^{y\left(\frac{\pi}{6}\right)} $ is $ 3 $.
Let $y=y(x)$ be the solution of the differential equation
$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x}{\left(1+x^2\right)^2} y=x \mathrm{e}^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0.$
Then the area enclosed by the curve $f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y-x=4$ is ________.
Explanation:
$\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{1+x^2}} ; y(0)=0$
I.F. of linear differential equation,
$\begin{aligned} & \text { I.F. }=e^{\int \frac{2 x}{\left(1+x^2\right)^2}} d x=e^{\left(\frac{-1}{1+x^2}\right)} \\ & \Rightarrow y\left(e^{\left(\frac{-1}{1+x^2}\right)}\right)=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\left(\frac{-1}{1+x^2}\right)} d x \\ & =\frac{x^2}{2}+c \\ & \Rightarrow y(0)=0 \Rightarrow 0\left(e^{-1}\right)=c \Rightarrow c=0 \\ & \Rightarrow y=\frac{e^{\frac{1}{1+x^2}} \cdot x^2}{2} \end{aligned}$
Area between curve $y e^{\left(\frac{-1}{1+x^2}\right)}=\frac{x^2}{2}$ and $y-x=4$
$\begin{aligned} & \Rightarrow 2(x+4)=x^2 \Rightarrow x^2-2 x-8=0 \\ & \Rightarrow(x-4)(x+2)=0 \\ & \int_{-2}^4\left[(x+4)-\frac{x^2}{2}\right] d x=\frac{x^2}{2}+4 x-\left.\frac{x^3}{6}\right|_{-2} ^4 \\ & =\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\ & =30-12=18 \end{aligned}$
Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^2 d x=d y, y(0)=-2$. Let the maximum and minimum values of the function $y=y(x)$ in $\left[0, \frac{\pi}{3}\right]$ be $\alpha$ and $\beta$, respectively. If $(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}$, then $\gamma+\delta$ equals _________.
Explanation:
$\begin{aligned} & \frac{d y}{d x}=(x+y+z)^2 \\ & \text { Put } x+y+z=t \\ & \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \\ & \text { Given DE } \Rightarrow \frac{d t}{d x}-1=t^2 \\ & \Rightarrow \frac{d t}{1+t^2}=d x \Rightarrow \tan ^{-1} t=x+c \\ & \Rightarrow x+y+z=\tan (x+c) \\ & \Rightarrow y(x)=\tan (x+c)-x-2 \\ & \because y(0)=-2 \Rightarrow-2=\tan c-0-2 \\ & \qquad \Rightarrow c=0 \\ & \Rightarrow y(x)=\tan x-x-2 \\ & \frac{d y}{d x}=\sec ^2 x-1 \geq 0 \end{aligned}$
$\Rightarrow y(x)$ is increasing if $x \in\left(0, \frac{\pi}{3}\right)$
$\begin{aligned} & \Rightarrow \alpha=y\left(\frac{\pi}{3}\right), \beta=y(0) \\ & \Rightarrow \alpha=-\frac{\pi}{3}-2+\sqrt{3} \text { and } \beta=-2 \end{aligned}$
Now, $(3 \alpha+\pi)^2+\beta^2=(6+3 \sqrt{3})^2+(-2)^2$
$=67-36 \sqrt{3}=y+\delta \sqrt{3}$.
$\Rightarrow \gamma+\delta=31$
Let the solution $y=y(x)$ of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}-y=1+4 \sin x$ satisfy $y(\pi)=1$. Then $y\left(\frac{\pi}{2}\right)+10$ is equal to __________.
Explanation:
$\begin{aligned} & \frac{d y}{d x}-y=1+4 \sin x \\ & \text { Integrating factor }=e^{-\int d x}=e^{-x} \end{aligned}$
Solution is $y e^{-x}=\int(1+4 \sin x) e^{-x} d x$
$\begin{aligned} & =-e^{-x}+2 \cdot e^{-x}(-\sin x-\cos x)+C \\ y(\pi) & =1 \Rightarrow C=0 \end{aligned}$
Hence $y(x)=-1-2(\sin x+\cos x)$
$y\left(\frac{\pi}{2}\right)+10=7$
Explanation:
Integrating factor $=\mathrm{e}^{\int-\frac{1}{y} d y}=\frac{1}{y}$
$\begin{aligned} & x \cdot \frac{1}{y}=\int \frac{1-y^2}{y^2} d y \\\\ & \frac{x}{y}=\frac{-1}{y}-y+c \\\\ & x=-1-y^2+c y\end{aligned}$
$\begin{aligned} & x(1)=1 \\\\ & 1=-1-1+c \Rightarrow c=3 \\\\ & x=-1-y^2+3 y \\\\ & 5 x(2)=5(-1-4+6) \\\\ & =5\end{aligned}$
Explanation:
$I.F=e^{-\int \frac{2}{t+1} d t}=e^{-2 \ln (t+1)}=\frac{1}{(t+1)^2}$
$\begin{aligned} & \frac{x}{(t+1)^2}=\int \frac{1}{(t+1)^2}(t+1)^3 d t+C\end{aligned}$
$\frac{x}{(t+1)^2}=\frac{t^2}{2}+t+C$
Now $x(0)=2$
$ \begin{aligned} & \Rightarrow C=2 \\\\ & \therefore x=\left(\frac{t^2}{2}+t+2\right)(t+1)^2 \\\\ & x(1)=\left(\frac{1}{2}+1+2\right)(1+1)^2 \\\\ & =\frac{7}{2} \times 4=14 \end{aligned} $
Let $y=y(x)$ be the solution of the differential equation
$\sec ^2 x d x+\left(e^{2 y} \tan ^2 x+\tan x\right) d y=0,0< x<\frac{\pi}{2}, y(\pi / 4)=0$.
If $y(\pi / 6)=\alpha$, then $e^{8 \alpha}$ is equal to ____________.
Explanation:
$\begin{aligned} & \sec ^2 x \frac{d x}{d y}+e^{2 y} \tan ^2 x+\tan x=0 \\ & \left(\text { Put } \tan x=t \Rightarrow \sec ^2 x \frac{d x}{d y}=\frac{d t}{d y}\right) \\ & \frac{d t}{d y}+e^{2 y} \times t^2+t=0 \\ & \frac{d t}{d y}+t=-t^2 \cdot e^{2 y} \\ & \frac{1}{t^2} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y} \\ & \left(\text { Put } \frac{1}{t}=u \frac{-1}{t^2} \frac{d t}{d y}=\frac{d u}{d y}\right) \\ & \frac{-d u}{d y}+u=-e^{2 y} \\ & \frac{d u}{d y}-u=e^{2 y} \\ & \text { I.F. }=e^{-\int d y}=e^{-y} \\ & u e^{-y}=\int e^{-y} \times e^{2 y} d y \\ & \frac{1}{\tan x} \times e^{-y}=e^y+c \\ & x=\frac{\pi}{4}, y=0, c=0 \\ & \begin{aligned} & \mathrm{x}=\frac{\pi}{6}, \quad \mathrm{y}=\alpha \\ & \sqrt{3} \mathrm{e}^{-\alpha}=\mathrm{e}^\alpha+0 \\ & \mathrm{e}^{2 \alpha}=\sqrt{3} \\ & \mathrm{e}^{8 \alpha}=9 \end{aligned} \end{aligned}$
Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y-y=Y^{\prime}(x)(X-x)$ and the co-ordinate axes, where $(x, y)$ is any point on the curve, is always $\frac{-y^2}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) \neq 0$. If $Y(1)=1$, then $12 Y(2)$ equals __________.
Explanation:
$\mathrm{A}=\frac{1}{2}\left(\frac{-\mathrm{y}}{\mathrm{Y}^{\prime}(\mathrm{x})}+\mathrm{x}\right)(\mathrm{y}-\mathrm{xY} / \mathrm{x})=\frac{-\mathrm{y}^2}{2 \mathrm{Y}^{\prime}(\mathrm{x})}+1$
$\Rightarrow\left(-y+x Y^{\prime}(x)\right)\left(y-x Y^{\prime}(x)\right)=-y^2+2 Y^{\prime}(x)$
$\begin{aligned} -y^2+x y Y^{\prime}(x)+x y Y^{\prime}(x) & -x^2\left[Y^{\prime}(x)\right]^2 \\ = & -y^2+2 Y^{\prime}(x) \end{aligned}$
$\begin{aligned} & 2 x y-x^2 Y^{\prime}(x)=2 \\ & \frac{d y}{d x}=\frac{2 x y-2}{x^2} \\ & \frac{d y}{d x}-\frac{2}{x} y=\frac{-2}{x^2} \\ & \text { I.F. }=e^{-2 \ln x}=\frac{1}{x^2} \\ & y \cdot \frac{1}{x^2}=\frac{2}{3} x^{-3}+c \\ & \text { Put } x=1, y=1 \\ & 1=\frac{2}{3}+c \Rightarrow c=\frac{1}{3} \\ & Y=\frac{2}{3} \cdot \frac{1}{X}+\frac{1}{3} X^2 \\ \Rightarrow \quad & 12 Y(2)=\frac{5}{3} \times 12=20 \end{aligned}$
Let $y=y(x)$ be the solution of the differential equation $\left(1-x^2\right) \mathrm{d} y=\left[x y+\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}\right] \mathrm{d} x, -1< x<1, y(0)=0$. If $y\left(\frac{1}{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}, \mathrm{m}$ and $\mathrm{n}$ are co-prime numbers, then $\mathrm{m}+\mathrm{n}$ is equal to __________.
Explanation:
$\begin{aligned} & \frac{d y}{d x}-\frac{x y}{1-x^2}=\frac{\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}}{1-x^2} \\ & \mathrm{IF}=e^{-\int \frac{x}{1-x^2} d x}=e^{+\frac{1}{2} \ln \left(1-x^2\right)}=\sqrt{1-x^2} \\ & y \sqrt{1-x^2}=\sqrt{3} \int\left(x^3+2\right) d x \\ & y \sqrt{1-x^2}=\sqrt{3}\left(\frac{x^4}{4}+2 x\right)+c \\ & \Rightarrow y(0)=0 \quad \therefore c=0\\ & y\left(\frac{1}{2}\right)=\frac{65}{32}=\frac{m}{n} \\ & m+n=97 \end{aligned}$
If the solution curve $y=y(x)$ of the differential equation $\left(1+y^2\right)\left(1+\log _{\mathrm{e}} x\right) d x+x d y=0, x > 0$ passes through the point $(1,1)$ and $y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$, then $\alpha+2 \beta$ is _________.
Explanation:
$\begin{aligned} & \int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \\ & \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C \end{aligned}$
Put $x=y=1$
$\begin{aligned} & \therefore C=\frac{\pi}{4} \\ & \Rightarrow \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4} \end{aligned}$
Put $x=e$
$\begin{aligned} & \Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}} \\ & \therefore \alpha=1, \beta=1 \\ & \Rightarrow \alpha+2 \beta=3 \end{aligned}$
If the solution curve, of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y-2}{x-y}$ passing through the point $(2,1)$ is $\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _{\mathrm{e}}\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log _{\mathrm{e}}|x-1|$, then $5 \beta+\alpha$ is equal to __________.
Explanation:
$\begin{aligned} & \frac{d y}{d x}=\frac{x+y-2}{x-y} \\ & \mathrm{x}=\mathrm{X}+\mathrm{h}, \mathrm{y}=\mathrm{Y}+\mathrm{k} \\ & \frac{d Y}{d X}=\frac{X+Y}{X-Y} \\ & \left.\begin{array}{l} \mathrm{h}+\mathrm{k}-2=0 \\ \mathrm{~h}-\mathrm{k}=0 \end{array}\right\} \mathrm{h}=\mathrm{k}=1 \\ & \mathrm{Y}=\mathrm{vX} \\ & v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^2}{1-v} \\ & \end{aligned}$
$\begin{aligned} & \frac{1-v}{1+v^2} d v=\frac{d X}{X} \\ & \tan ^{-1} v-\frac{1}{2} \ln \left(1+v^2\right)=\ln |X|+C \end{aligned}$
As curve is passing through $(2,1)$
$\begin{aligned} & \tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln |x-1| \\ & \therefore \alpha=1 \text { and } \beta=2 \\ & \Rightarrow 5 \beta+\alpha=11 \end{aligned}$
$(2 x+3 y-2) \mathrm{d} x+(4 x+6 y-7) \mathrm{d} y=0, y(0)=3$, is
$\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6$, then $\alpha+2 \beta+3 \gamma$ is equal to ____________.
Explanation:
$\begin{array}{ll} 2 x+3 y-2=t & 4 x+6 y-4=2 t \\ 2+3 \frac{d y}{d x}=\frac{d t}{d x} & 4 x+6 y-7=2 t-3 \end{array}$
$\begin{aligned} & \frac{d y}{d x}=\frac{-(2 x+3 y-2)}{4 x+6 y-7} \\ & \frac{d t}{d x}=\frac{-3 t+4 t-6}{2 t-3}=\frac{t-6}{2 t-3} \\ & \int \frac{2 t-3}{t-6} d t=\int d x \\ & \int\left(\frac{2 t-12}{t-6}+\frac{9}{t-6}\right) \cdot d t=x \\ & 2 t+9 \ln (t-6)=x+c \\ & 2(2 x+3 y-2)+9 \ln (2 x+3 y-8)=x+c \\ & x=0, y=3 \\ & c=14 \\ & 4 x+6 y-4+9 \ln (2 x+3 y-8)=x+14 \\ & x+2 y+3 \ln (2 x+3 y-8)=6 \\ & \alpha=1, \beta=2, \gamma=8 \\ & \alpha+2 \beta+3 \gamma=1+4+24=29 \end{aligned}$
$2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0, y>-1$
with $x\left(e^{4}-2\right)=1$. Then $x\left(e^{9}-2\right)$ is equal to :
Let $y=y_{1}(x)$ and $y=y_{2}(x)$ be the solution curves of the differential equation $\frac{d y}{d x}=y+7$ with initial conditions $y_{1}(0)=0$ and $y_{2}(0)=1$ respectively. Then the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ intersect at
Let $y=y(x), y > 0$, be a solution curve of the differential equation $\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x$. If $y(0)=1$ and $y(2 \sqrt{2})=\beta$, then
Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0$. If $y(1)=2$, then $y(2)$ is equal to :
Let $y=y(x)$ be a solution curve of the differential equation.
$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$.
If the line $x=1$ intersects the curve $y=y(x)$ at $y=2$ and the line $x=2$ intersects the curve $y=y(x)$ at $y=\alpha$, then a value of $\alpha$ is :
Let $f$ be a differentiable function such that ${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt} $, $f(1) = {2 \over 3}$. Then $18f(3)$ is equal to :
If the solution curve $f(x, y)=0$ of the differential equation
$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$,
passes through the points $(1,0)$ and $(\alpha, 2)$, then $\alpha^{\alpha}$ is equal to :
Let $\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$ be the solution of the differential equation $2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2}$. Then $\alpha+\beta-\gamma$ equals :
The area enclosed by the closed curve $\mathrm{C}$ given by the differential equation
$\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $\mathrm{C}$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $\mathrm{C}$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is :
If $y=y(x)$ is the solution curve of the differential equation
$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$, then $y\left(\frac{\pi}{6}\right)$ is equal to
$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$
such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to :
Let a differentiable function $f$ satisfy $f(x)+\int_\limits{3}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$. Then $12 f(8)$ is equal to :
$\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is :
Let the solution curve $y=y(x)$ of the differential equation
$
\frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text { pass through the origin. Then } y(1) \text { is equal to : }
$
Let $y=y(x)$ be the solution of the differential equation $x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1)$. If $y(2) = 2$, then $y(e)$ is equal to
Let $y=f(x)$ be the solution of the differential equation $y(x+1)dx-x^2dy=0,y(1)=e$. Then $\mathop {\lim }\limits_{x \to {0^ + }} f(x)$ is equal to
Let $y=y(t)$ be a solution of the differential equation ${{dy} \over {dt}} + \alpha y = \gamma {e^{ - \beta t}}$ where, $\alpha > 0,\beta > 0$ and $\gamma > 0$. Then $\mathop {\lim }\limits_{t \to \infty } y(t)$
Let $y = y(x)$ be the solution curve of the differential equation ${{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3$. Then ${{{y^2}(x)} \over 9}$ is equal to :
Let $y=y(x)$ be the solution of the differential equation $(x^2-3y^2)dx+3xy~dy=0,y(1)=1$. Then $6y^2(e)$ is equal to
Let $y = y(x)$ be the solution of the differential equation ${x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}$. Then y (1) is equal to
If $y=y(x)$ is the solution of the differential equation
$\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x > 1$ such that
$y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to }$ :
Explanation:
We can solve the given differential equation using an integrating factor.
The integrating factor is given by :
$
\mu(x) = e^{\int \frac{4x}{x^2 - 1} dx} = e^{2\ln(x^2 - 1)} = (x^2 - 1)^2
$
Multiplying both sides of the differential equation by $\mu(x)$, we get :
$(x^2-1)^2 \frac{d y}{d x} + 4x y = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$
We can rewrite the left-hand side using the product rule:
$\frac{d}{dx} \left((x^2-1)^2 y\right) = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$
Integrating both sides with respect to $x$, we get:
$(x^2-1)^2 y = \int \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}} dx = \sqrt{x^2-1}+2 \ln\left|x+\sqrt{x^2-1}\right|+C$
where $C$ is the constant of integration. Using the initial condition $y(2) = \frac{2}{9} \log_e(2+\sqrt{3})$, we can solve for $C$:
At $x=2$,$9 \cdot \frac{2}{9} \ln (2+\sqrt{3})=2 \ln (2+\sqrt{3})+\sqrt{3}+C$
$C=-\sqrt{3}$
At $x=\sqrt{2}$
$y(\sqrt{2})=2 \ln (1+\sqrt{2})+1-\sqrt{3}$
$\beta=1, \alpha=2, \gamma=3$
$ \Rightarrow \alpha \beta \gamma=6 $
Let the tangent at any point P on a curve passing through the points (1, 1) and $\left(\frac{1}{10}, 100\right)$, intersect positive $x$-axis and $y$-axis at the points A and B respectively. If $\mathrm{PA}: \mathrm{PB}=1: k$ and $y=y(x)$ is the solution of the differential equation $e^{\frac{d y}{d x}}=k x+\frac{k}{2}, y(0)=k$, then $4 y(1)-6 \log _{\mathrm{e}} 3$ is equal to ____________.
Explanation:
Whose slope is $\frac{d y}{d x}$ is
$ \begin{aligned} &\mathrm{Y}-y=\frac{d y}{d x}(\mathrm{X}-x)\\\\ &\begin{aligned} & \text { Putting } Y=0 \Rightarrow X=x-\frac{y}{\left(\frac{d y}{d x}\right)} \mathrm{t} \\\\ & \Rightarrow \alpha=x-\frac{y}{\left(\frac{d y}{d x}\right)} \end{aligned} \end{aligned} $
and putting $\mathrm{X}=0 \Rightarrow \mathrm{Y}=y-x \frac{d y}{d x}$
$ \Rightarrow \beta=y-x \frac{d y}{d x} $
$\because \mathrm{P}$ divides $\mathrm{AB}$ in $1: k$
$ x=\frac{k \alpha+0}{k+1} \text { and } y=\frac{k \times 0+\beta}{k+1} $
$ \begin{aligned} & \Rightarrow x(k+1)=k\left(x-\frac{y}{\frac{d y}{d x}}\right) \\\\ & \Rightarrow x k+x=x k-\frac{y k}{\frac{d y}{d x}} \\\\ &\Rightarrow x \frac{d y}{d x}=-y k \end{aligned} $
$ \begin{aligned} & \text { or } \int \frac{d y}{y}=-k \times \int \frac{1}{x} d x \\\\ & \Rightarrow \log y=-k \log x+\log \mathrm{C} \\\\ & \text { or } \log y \times x^k=\log \mathrm{C} \\\\ & \Rightarrow y x^k=\mathrm{C} \\\\ & \text { putting } x=1, y=1 \Rightarrow c=1 \\\\ & \text { so } y x^k=1 \end{aligned} $
Putting $x=\frac{1}{10}, y=100 \Rightarrow 100 \times\left(\frac{1}{100}\right)^k=1 \Rightarrow k=2$
so $y x^2=1$ or $y=\frac{1}{x^2}$
Now $e^{\frac{d y}{d x}}=k x+\frac{k}{2}$
$ \Rightarrow \frac{d y}{d x}=\log _e\left(k x+\frac{k}{2}\right)=\log _e(2 x+1) $
On integrating
$ \begin{aligned} & y=\int 1 \cdot \log _e(2 x+1) d x \\\\ & =x \log _e(2 x+1)-\int \frac{1 \times 2}{2 x+1} \times x d x \\\\ & =x \log _e(2 x+1)-\int 1-\frac{1}{2 x+1} d x \\\\ & y=x \log _e(2 x+1)-x+\frac{1}{2} \log _e(2 x+1)+c \end{aligned} $
Put $x=0, y=k=2 \Rightarrow c=2$ putting $x=1$
$ \begin{aligned} & y(1)=\log _e 3-1+\frac{1}{2} \log _e 3+2=\frac{3}{2} \log _e 3-1+2 \\\\ & \Rightarrow 4 y(1)=6 \log _e 3+4 \\\\ & \Rightarrow 4 y(1)-6 \log _e 3= 4 \end{aligned} $
Let the solution curve $x=x(y), 0 < y < \frac{\pi}{2}$, of the differential equation $\left(\log _{e}(\cos y)\right)^{2} \cos y \mathrm{~d} x-\left(1+3 x \log _{e}(\cos y)\right) \sin \mathrm{y} d y=0$ satisfy $x\left(\frac{\pi}{3}\right)=\frac{1}{2 \log _{e} 2}$. If $x\left(\frac{\pi}{6}\right)=\frac{1}{\log _{e} m-\log _{e} n}$, where $m$ and $n$ are coprime, then $m n$ is equal to __________.
Explanation:
Which is a linear differential equation.
$ \text { I.F. }=e^{-\int \frac{3 \tan y}{\ln \cos y} d y}=(\ln \cos y)^3 $
So, the solution is :
$ \begin{aligned} & x \times(\ln \cos y)^3=\int\left((\ln \cos y)^3 \times \frac{\tan y}{(\ln \cos y)^2}\right) d y \\\\ & x \times(\ln \cos y)^3=\frac{-(\ln \cos y)^2}{2}+C \end{aligned} $
$ \text { At } y=\frac{\pi}{3} \text {, } $
$ \begin{aligned} & \frac{1}{2 \ln 2} \times\left(\ln \left(\frac{1}{2}\right)\right)^3=-\frac{\left(\ln \left(\frac{1}{2}\right)\right)^2}{2}+C \\\\ & \Rightarrow C=0 \end{aligned} $
$ \begin{aligned} & \text { So, } x \times \ln ^3 \cos y=\frac{-\ln ^2 \cos y}{2} \\\\ & \text { At } y=\frac{\pi}{6}, x \times\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^3=-\frac{1}{2}\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^2 \\\\ & \Rightarrow x=-\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} \end{aligned} $
$ \begin{aligned} & =-\frac{1}{2[\ln \sqrt{3}-\ln 2]}=\frac{-1}{2\left[\frac{1}{2} \ln 3-\ln 2\right]} \\\\ & =\frac{-1}{2\left[\frac{\ln 3-\ln 4}{2}\right]}=\frac{1}{\ln 4-\ln 3} \\\\ & \Rightarrow m=4, n=3 \\\\ & \Rightarrow m n=12 \end{aligned} $
If the solution curve of the differential equation $\left(y-2 \log _{e} x\right) d x+\left(x \log _{e} x^{2}\right) d y=0, x > 1$ passes through the points $\left(e, \frac{4}{3}\right)$ and $\left(e^{4}, \alpha\right)$, then $\alpha$ is equal to ____________.
Explanation:
$ \begin{aligned} & (y-2 \log x) d x+\left(x \log x^2\right) d y=0 \\\\ & \Rightarrow \frac{d y}{d x}=\frac{(2 \log x-y)}{2 x \log x} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{y}{2 x \log x}=\frac{1}{x} \end{aligned} $
It is a linear differential equation.
$ \therefore \text { I.F. }=e^{\int \frac{1}{2 x \log x} d x} $
Put $\log x=t \Rightarrow \frac{1}{x} d x=d t$
$ \therefore \text { I.F. }=e^{\int \frac{1}{2 t} d t}=e^{\log (t)^{\frac{1}{2}}}=\sqrt{t}=\sqrt{\log x} $
So, required solution is,
$ \begin{aligned} & y \sqrt{\log x}=\int \frac{\sqrt{\log x}}{x} d x \\\\ & \log x=v \Rightarrow \frac{1}{x} d x=d v \\\\ & \Rightarrow y \sqrt{\log x}=\int \sqrt{v} d v+C \\\\ & \Rightarrow y \sqrt{\log x}=\frac{2 v^{3 / 2}}{3}+C \\\\ & \Rightarrow y \sqrt{\log x}=\frac{2}{3}(\log x)^{3 / 2}+C \end{aligned} $
Now, this curve passes through $\left(e, \frac{4}{3}\right)$ and $\left(e^4, \alpha\right)$
$ \begin{aligned} & \therefore \frac{4}{3} \sqrt{\log e}=\frac{2}{3}(\log e)^{3 / 2}+C \\\\ & \Rightarrow C=\frac{4}{3}-\frac{2}{3}=\frac{2}{3} \end{aligned} $
Also, $\alpha \sqrt{\log e^4}=\frac{2}{3}\left(\log e^4\right)^{3 / 2}+\frac{2}{3}$
$ \begin{aligned} & \Rightarrow 2 \alpha=\frac{2}{3} \times(4)^{3 / 2}+\frac{2}{3}=\frac{16}{3}+\frac{2}{3}=\frac{18}{3} \\\\ & \Rightarrow \alpha=3 \end{aligned} $
Let $y=y(x)$ be a solution of the differential equation $(x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2}$. If $\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3}$, then $\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|$ is equal to ____________.
Explanation:
$ \begin{aligned} & (x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{x y \sin x+y \cos x-1}{x \cos x}=0 \\\\ & \Rightarrow \frac{d y}{d x}+\left(\frac{x y \sin x+y \cos x}{x \cos x}\right)=\frac{1}{x \cos x} \\\\ & \Rightarrow \frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{1}{x \cos x} \end{aligned} $
Which is linear differential equation in the form of
$ \begin{aligned} & \frac{d y}{d x}+P y=Q \\\\ & \therefore \mathrm{IF}=e^{\int\left(\tan x+\frac{1}{x}\right) d x}=e^{(\log \sec x+\log x)}=e^{\log (x \sec x)}=x \sec x \end{aligned} $
$\therefore$ The general solution of the given differential equation
$ \begin{array}{rlrl} y \cdot \mathrm{IF} =\int(Q \times \mathrm{IF}) d x+c \\\\ \Rightarrow y(x \sec x) =\int\left(\frac{1}{x \cos x} x \sec x\right) d x+c \\\\ \Rightarrow x y \sec x =\int \sec ^2 x d x+c \\\\ \Rightarrow x y \sec x =\tan x+c ........(i) \end{array} $
Since, $\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3} \Rightarrow y\left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{\pi}$
$ \begin{aligned} & \therefore \frac{\pi}{3}\left(\frac{3 \sqrt{3}}{\pi}\right) \sec \left(\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{3}\right)+c \\\\ & \Rightarrow \sqrt{3}(2)=\sqrt{3}+c \\\\ & \begin{aligned} \Rightarrow & c=\sqrt{3} \end{aligned} \end{aligned} $
On putting the value of $c$ in Eq. (i), we get
$ \begin{aligned} &x y \sec x =\tan x+\sqrt{3} \\\\ &\Rightarrow y =\frac{1}{x}(\sin x+\sqrt{3} \cos x) \\\\ & y^{\prime} =\frac{1}{x}(\cos x-\sqrt{3} \sin x)-\frac{1}{x^2}(\sin x+\sqrt{3} \cos x) \end{aligned} $
$\begin{aligned} & \text { and } y^{\prime \prime}=\frac{1}{x}(-\sin x-\sqrt{3} \cos x)-\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) -\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) \\ & -\frac{2}{x^3}(\cos x-\sqrt{3} \sin x) \\\\ & \therefore\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|=|-2|=2 \\\\ & \end{aligned}$
If the solution curve of the differential equation $\frac{d y}{d x}=\frac{x+y-2}{x-y}$ passes through the points $(2,1)$ and $(\mathrm{k}+1,2), \mathrm{k}>0$, then
Let $y=y(x)$ be the solution curve of the differential equation $ \frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$, which passes through the point $(0,1)$. Then $y(1)$ is equal to :
Let the solution curve $y=y(x)$ of the differential equation $\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1$ pass through the point $\left(0, \frac{\pi}{2}\right)$. Then, $\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)$ is equal to :
Let $y=y(x)$ be the solution curve of the differential equation $ \frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}$, $x >1$ passing through the point $\left(2, \sqrt{\frac{1}{3}}\right)$. Then $\sqrt{7}\, y(8)$ is equal to :
The differential equation of the family of circles passing through the points $(0,2)$ and $(0,-2)$ is :
Let the solution curve of the differential equation $x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$, intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$. Then the value of $\alpha$ is :
If $y=y(x), x \in(0, \pi / 2)$ be the solution curve of the differential equation
$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$,
with $y(\pi / 4)=\mathrm{e}^{-\pi}$, then $y(\pi / 6)$ is equal to :
Let $y=y_{1}(x)$ and $y=y_{2}(x)$ be two distinct solutions of the differential equation $\frac{d y}{d x}=x+y$, with $y_{1}(0)=0$ and $y_{2}(0)=1$ respectively. Then, the number of points of intersection of $y=y_{1}(x)$ and $y=y_{2}(x)$ is