JEE MAIN Differential Equation Previous Year Questions (PYQs) – Page 11 of 16

JEE MAIN Differential Equation Previous Year Questions (PYQs) – Page 11 of 16

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🎓 JEE MAIN📅 Year: 2022📚 Mathematics🏷 Differential Equation

Let x = x(y) be the solution of the differential equation $2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$ such that x(1) = 0. Then, x(e) is equal to :

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🎓 JEE MAIN📅 Year: 2019📚 Mathematics🏷 Differential Equation

The solution of the differential equation $x\dfrac{dy}{dx} + 2y = x^2 \ (x \ne 0)$ with $y(1) = 1$, is:

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🎓 JEE MAIN📅 Year: 2022📚 Mathematics🏷 Differential Equation

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $\tan x(\cos x - y)$. If the curve passes through the point $\left( {{\pi \over 4},0} \right)$, then the value of $\int\limits_0^{\pi /2} {y\,dx} $ is equal to :

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🎓 JEE MAIN📅 Year: 2024📚 Mathematics🏷 Differential Equation

The solution curve of the differential equation $y\dfrac{dx}{dy}=x(\log_e x-\log_e y+1),\ x>0,\ y>0,$ passing through the point $(e,1)$ is:

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🎓 JEE MAIN📅 Year: 2024📚 Mathematics🏷 Differential Equation

Let $\displaystyle \int_{0}^{x}\sqrt{1-\big(y'(t)\big)^{2}},dt=\int_{0}^{x}y(t),dt,\ 0\le x\le 3,\ y\ge0,\ y(0)=0$. Then at $x=2$, $,y''+y+1$ is equal to:

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🎓 JEE MAIN📅 Year: 2019📚 Mathematics🏷 Differential Equation

If $\cos x{{dy} \over {dx}} - y\sin x = 6x$, (0 < x < ${\pi \over 2}$)
and $y\left( {{\pi \over 3}} \right)$ = 0 then $y\left( {{\pi \over 6}} \right)$ is equal to :

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🎓 JEE MAIN📅 Year: 2016📚 Mathematics🏷 Differential Equation

If a curve $y = f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation $ y(1+xy),dx = x,dy $, then $ f\left(-\dfrac{1}{2}\right) $ is equal to:

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🎓 JEE MAIN📅 Year: 2025📚 Mathematics🏷 Differential Equation

Let $x=x(y)$ be the solution of the differential equation $2(y+2)\log_e(y+2)\,dx+\big(x+4-2\log_e(y+2)\big)\,dy=0,\quad y>-1$ with $x\big(e^{4}-2\big)=1$. Then $x\big(e^{9}-2\big)$ is equal to:

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🎓 JEE MAIN📅 Year: 2021📚 Mathematics🏷 Differential Equation

If ${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$, y(0) = 1, then y(1) is equal to :

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🎓 JEE MAIN📅 Year: 2022📚 Mathematics🏷 Differential Equation

Let the solution curve of the differential equation

$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $, $y(1) = 3$ be $y = y(x)$. Then y(2) is equal to:


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