JEE MAIN Differential Equation Previous Year Questions (PYQs) – Page 1 of 16
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Let the population of rabbits surviving at time $t$ be governed by the differential equation
$\dfrac{dp(t)}{dt}=\dfrac{1}{2}p(t)-200$. If $p(0)=100$, then $p(t)$ equals:
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Solution
The curve amongst the family of curves represented by the differential equation
$(x^2 - y^2)dx + 2xy\,dy = 0$
which passes through $(1, 1)$ is :
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Solution
The area enclosed by the closed curve $\mathcal{C}$ given by the differential equation
$\dfrac{dy}{dx}+\dfrac{x+a}{\,y-2\,}=0,\quad y(1)=0$
is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $\mathcal{C}$ with the $y$-axis. If the normals at $P$ and $Q$ on $\mathcal{C}$ intersect the $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is:
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Solution
Let y = y(x) be the solution of the differential equation xdy = (y + x3 cosx)dx with y($\pi$) = 0, then $y\left( {{\pi \over 2}} \right)$ is equal to :
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Solution
If $y=y(x)$ is the solution curve of the differential equation $\dfrac{dy}{dx}+y\tan x=x\sec x,\ 0\le x\le \dfrac{\pi}{3},\ y(0)=1$, then $y\!\left(\dfrac{\pi}{6}\right)$ is equal to:
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Solution
The general solution of the differential equation (y2
– x3)dx – xydy = 0 (x $ \ne $ 0) is :
(where c is a constant of integration)
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Solution
Let $y = y(x)$ be the solution of differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $y(0) = -\frac{7}{4}$. Then $y\left(\frac{\pi}{6}\right)$ is equal to:
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Solution
$\frac{dy}{dx} - 2y\cos x = 2\cos x + 3\sin x \cos x$
$I.F. = e^{-2\sin x}$
$e^{-2\sin x}y = \int e^{-2\sin x}(3\sin x \cos x + 2\cos x),dx$
$= e^{-2\sin x}\left(-\frac{3}{2}\sin x - \frac{7}{4}\right) + C$
$\Rightarrow y = -\frac{3}{2}\sin x - \frac{7}{4} + Ce^{2\sin x}$
$y(0) = -\frac{7}{4} \Rightarrow C = 0$
$y\left(\frac{\pi}{6}\right) = -\frac{3}{2}\cdot\frac{1}{2} - \frac{7}{4} = -\frac{5}{2}$
If a curve y=y(x) passes through the point $\left(1,\dfrac{\pi}{2}\right)$ and satisfies the differential equation $(7x^{4}\cot y-e^{x}\csc y),\dfrac{dx}{dy}=x^{5},\ x\ge1$, then at $x=2$, the value of $\cos y$ is:
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Solution
Given:
$(7x^{4}\cot y - e^{x}\csc y)\frac{dx}{dy}=x^{5}$
⇒ $\frac{dy}{dx}=\frac{7x^{4}\cot y - e^{x}\csc y}{x^{5}}$
Multiply by $\sin y$:
$\sin y\frac{dy}{dx}=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
Now note:
$\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}$
So,
$-\frac{d}{dx}(\cos y)=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
⇒ $\frac{d}{dx}(\cos y)+\frac{7}{x}\cos y=\frac{e^{x}}{x^{5}}$
This is linear in $\cos y$
IF $=e^{\int \frac{7}{x}dx}=x^{7}$
$\frac{d}{dx}(x^{7}\cos y)=x^{2}e^{x}$
Integrate:
$x^{7}\cos y=\int x^{2}e^{x}dx$
$\int x^{2}e^{x}dx=e^{x}(x^{2}-2x+2)$
So,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)+C$
Use $(1,\frac{\pi}{2})$ ⇒ $\cos y=0$
$0=e(1-2+2)+C=e+C$ ⇒ $C=-e$
Thus,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)-e$
At $x=2$:
$2^{7}\cos y=e^{2}(4-4+2)-e=2e^{2}-e$
$\cos y=\frac{2e^{2}-e}{128}=\frac{e(2e-1)}{128}$
$\boxed{\frac{e(2e-1)}{128}}$
⇒ $\frac{dy}{dx}=\frac{7x^{4}\cot y - e^{x}\csc y}{x^{5}}$
Multiply by $\sin y$:
$\sin y\frac{dy}{dx}=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
Now note:
$\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}$
So,
$-\frac{d}{dx}(\cos y)=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$
⇒ $\frac{d}{dx}(\cos y)+\frac{7}{x}\cos y=\frac{e^{x}}{x^{5}}$
This is linear in $\cos y$
IF $=e^{\int \frac{7}{x}dx}=x^{7}$
$\frac{d}{dx}(x^{7}\cos y)=x^{2}e^{x}$
Integrate:
$x^{7}\cos y=\int x^{2}e^{x}dx$
$\int x^{2}e^{x}dx=e^{x}(x^{2}-2x+2)$
So,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)+C$
Use $(1,\frac{\pi}{2})$ ⇒ $\cos y=0$
$0=e(1-2+2)+C=e+C$ ⇒ $C=-e$
Thus,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)-e$
At $x=2$:
$2^{7}\cos y=e^{2}(4-4+2)-e=2e^{2}-e$
$\cos y=\frac{2e^{2}-e}{128}=\frac{e(2e-1)}{128}$
$\boxed{\frac{e(2e-1)}{128}}$
Let the solution curve $y=f(x)$ of the differential equation
$\dfrac{dy}{dx}+\dfrac{xy}{x^{2}-1}=\dfrac{x^{4}+2x}{\sqrt{1-x^{2}}}$, $x\in(-1,1)$,
pass through the origin. Then $\displaystyle \int_{-\sqrt{3}/2}^{\sqrt{3}/2} f(x)\,dx$ is equal to:
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Solution
Let $y=y(x)$ be the solution of the differential equation
$(x^{2}+4)^{2}\,dy+(2x^{3}y+8xy-2)\,dx=0.$
If $y(0)=0$, then $y(2)$ is equal to:
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Solution
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