Differential Equations · JEE Main previous year questions

2026 10 questions

1. Q252026 · 6 Apr 2026 · Shift 1· Numerical answer

   Let $y=y(x)$ be the solution of the differential equation $\left(x^2-x\sqrt{x^2-1}\right)dy+\left(y\left(x-\sqrt{x^2-1}\right)-x\right)dx=0$, $x\ge 1$. If $y(1)=1$, then the greatest integer less than $y(\sqrt{5})$ is ____.

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   Answer: 3

2. Q252026 · 5 Apr 2026 · Shift 2· Numerical answer

   Let $y=y(x)$ be the solution of the differential equation $$(\tan x{)}^{1/2}\mathrm{d}y=\left({\sec }^{3}x-(\tan x{)}^{3/2}y\right)\mathrm{d}x,0<x<\frac{\pi }{2},y\left(\frac{\pi }{4}\right)=\frac{6\sqrt{2}}{5}$$ If $y\left(\frac{\pi }{3}\right)=\frac{4}{5}\alpha$, then ${\alpha }^4$ equals

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   Answer: 48

3. Q202026 · 4 Apr 2026 · Shift 1

   Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(1+x+x^2)(1-y+y^2)$, $y(0)=\frac{1}{2}$. Then $(2y(1)-1)$ is equal to

   1. A.$\sqrt{3}\tan \left(\frac{11\sqrt{3}}{6}\right)$
   2. B.$\frac{\sqrt{3}}{2}\tan \left(\frac{11\sqrt{3}}{12}\right)$
   3. C.$\sqrt{3}\tan \left(\frac{11\sqrt{3}}{12}\right)$
   4. D.$\frac{\sqrt{3}}{2}\tan \left(\frac{11\sqrt{3}}{6}\right)$
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   Answer: C

4. Q192026 · 4 Apr 2026 · Shift 2

   Let $y=y(x)$ be the solution of the differential equation: $$\frac{\mathrm{d}y}{\mathrm{d}x}+\left(\frac{6x^2+\left(3x^2+2x^3+4\right)e^{-2x}}{(x^3+2)\left(2+e^{-2x}\right)}\right)y=2+e^{-2x},$$ $x\in (-1,2)$, satisfying $y(0)=\frac{3}{2}$. If $y(1)=\alpha (2+e^{-2})$, then $\alpha$ is equal to:

   1. A.$\frac{13}{8}$
   2. B.$\frac{6}{13}$
   3. C.$\frac{12}{13}$
   4. D.$\frac{13}{12}$
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   Answer: D

5. Q172026 · 2 Apr 2026 · Shift 1

   If the curve $y=f(x)$ passes through the point $(1,\mathrm{e})$ and satisfies the differential equation $\mathrm{d}y=y\left(2+{\log }_{\mathrm{e}}x\right)\mathrm{d}x,x>0$, then $f(\mathrm{e})$ is equal to :

   1. A.${\mathrm{e}}^{\mathrm{e}}$
   2. B.${\mathrm{e}}^{{\mathrm{e}}^2}$
   3. C.${\mathrm{e}}^{2\mathrm{e}}$
   4. D.${\mathrm{e}}^{2^{\mathrm{e}}}$
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   Answer: C

6. Q202026 · 2 Apr 2026 · Shift 1

   Let $y=y(x)$ be the solution curve of the differential equation $(1+\sin x)\frac{\mathrm{d}y}{\mathrm{d}x}+(y+1)\cos x=0,y(0)=0$. If the curve $y=y(x)$ passes through the point $\left(\alpha ,\frac{-1}{2}\right)$, then a value of $\alpha$ is :

   1. A.$\frac{\pi }{6}$
   2. B.$\frac{\pi }{4}$
   3. C.$\frac{\pi }{3}$
   4. D.$\frac{\pi }{2}$
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   Answer: D

7. Q172026 · 28 Jan 2026 · Shift 1

   Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-\sin 2y=x^3\left(2-x^3\right){\cos }^{2}y,x\ne 0$. If $y(2)=0$, then $\tan (y(1))$ is equal to

   1. A.$-\frac{7}{4}$
   2. B.$-\frac{3}{4}$
   3. C.$\frac{3}{4}$
   4. D.$\frac{7}{4}$
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   Answer: D

8. Q22026 · 28 Jan 2026 · Shift 2

   Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-y=x^2\cot x,x\in (0,\pi )$. If $y\left(\frac{\pi }{2}\right)=\frac{\pi }{2}$, then $6y\left(\frac{\pi }{6}\right)-8y\left(\frac{\pi }{4}\right)$ is equal to :

   1. A.$-3\pi$
   2. B.$3\pi$
   3. C.$-\pi$
   4. D.$\pi$
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   Answer: C

9. Q222026 · 23 Jan 2026 · Shift 2· Numerical answer

   If the solution curve $y=f(x)$ of the differential equation $$(x^2-4)y^{\prime }-2xy+2x(4-x^2{)}^2=0,x>2,$$ passes through the point $(3,15)$, then the local maximum value of $f$ is _____.

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   Answer: 16

10. Q72026 · 21 Jan 2026 · Shift 1

    Let $y=y(x)$ be the solution curve of the differential equation $(1+x^2)dy+(y-{\tan }^{-1}x)dx=0,y(0)=1$. Then the value of $y(1)$ is :

    1. A.$\frac{4}{{\mathrm{e}}^{\pi /4}}-\frac{\pi }{2}-1$
    2. B.$\frac{2}{{\mathrm{e}}^{\pi /4}}+\frac{\pi }{4}-1$
    3. C.$\frac{4}{{\mathrm{e}}^{\pi /4}}+\frac{\pi }{2}-1$
    4. D.$\frac{2}{{\mathrm{e}}^{\pi /4}}-\frac{\pi }{4}-1$
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    Answer: B