JEE MAIN Differential Equation Previous Year Questions (PYQs) – Page 12 of 16
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If $\log_{e} y = 3\sin^{-1}x$, then $,(1-x^{2})y''-xy',$ at $x=\dfrac{1}{2}$ is equal to:
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Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_1^1 f(\mathrm{t}) \mathrm{dt}=5 x f(x)-x^5-9$ for all $x \geqslant 1$, then the value of $f(3)$ is :
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Let $y=y(x)$ be the solution of the differential equation
$$x\log_e x \,\frac{dy}{dx}+y=x^2\log_e x,\quad (x>1).$$
If $y(2)=2$, then $y(e)$ is equal to:
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If x3dy + xy dx = x2dy + 2y dx; y(2) = e and x > 1, then y(4) is equal to :
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Let $ y = y(x) $ be the solution of the differential equation
$ x \frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y,; x \ne 0 $.
If $ y(2) = x $, then $ \tan(y(1)) $ is equal to
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Solution
$ x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y $
$ \sec^2 y \frac{dy}{dx} - 2\tan y \cdot \frac{1}{x} = x^2(2 - x^3) $
$ \tan y = t \Rightarrow \sec^2 y \frac{dy}{dx} = \frac{dt}{dx} $
$ \frac{dt}{dx} - \frac{2t}{x} = x^2(2 - x^3) \quad (\text{LDE}) $
I.F. $ = e^{\int -\frac{2}{x}dx} = e^{-2\ln x} = \frac{1}{x^2} $
$ \therefore \frac{t}{x^2} = \int \frac{1}{x^2} x^2(2 - x^3)dx + C $
$ \frac{\tan y}{x^2} = 2x - \frac{x^4}{4} + C $
$ y(2) = 0 \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0 $
$ \tan y = 2x^3 - \frac{1}{4}x^6 $
$ x = 1 \Rightarrow \tan y = 2 - \frac{1}{4} = \frac{7}{4} $
If ${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$, y(0) = 0, then for y = 1, the value of x lies in the interval :
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Solution
If $y{{dy} \over {dx}} = x\left[ {{{{y^2}} \over {{x^2}}} + {{\phi \left( {{{{y^2}} \over {{x^2}}}} \right)} \over {\phi '\left( {{{{y^2}} \over {{x^2}}}} \right)}}} \right]$, x > 0, $\phi$ > 0, and y(1) = $-$1, then $\phi \left( {{{{y^2}} \over 4}} \right)$ is equal to :
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The temperature $T(t)$ of a body at time $t=0$ is $160^\circ\!F$ and it decreases
continuously as per the differential equation $\dfrac{dT}{dt}=-K(T-80)$,
where $K$ is a positive constant. If $T(15)=120^\circ\!F$, then $T(45)$ is:
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Solution
Let y = y(x) be the solution of the differential equation $x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$, $ - 1 \le x \le 1$, $y\left( {{1 \over 2}} \right) = {\pi \over 6}$. Then the area of the region bounded by the curves x = 0, $x = {1 \over {\sqrt 2 }}$ and y = y(x) in the upper half plane is :
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If $y=y(x)$ is the solution of the differential equation
$x\dfrac{dy}{dx}+2y=x^{2}$, satisfying $y(1)=1$, then $y\!\left(\dfrac{1}{2}\right)$ is equal to:
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Solution
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