Differential Equations · JEE Main previous year questions

2020 7 questions

1. Q142020 · 6 Sept 2020 · Shift 2

   If $y=\left(\frac{2}{\pi }x-1\right)\mathrm{cosec}x$ is the solution of the differential equation, $$\frac{dy}{dx}+p(x)y=\frac{2}{\pi }\mathrm{cosec}x,$$ $0<x<\frac{\pi }{2}$, then the function $p(x)$ is equal to :

   1. A.$\cot x$
   2. B.$\sec x$
   3. C.$\tan x$
   4. D.$\mathrm{cosec}x$
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   Answer: A

2. Q142020 · 3 Sept 2020 · Shift 1

   The solution curve of the differential equation, $(1+e^{-x})(1+y^2)\frac{dy}{dx}=y^2$, which passes through the point $(0,1)$, is :

   1. A.$y^2+1=y\left({\log }_e\left(\frac{1+e^{-x}}{2}\right)+2\right)$
   2. B.$y^2+1=y\left({\log }_e\left(\frac{1+e^x}{2}\right)+2\right)$
   3. C.$y^2=1+y{\log }_e\left(\frac{1+e^{-x}}{2}\right)$
   4. D.$y^2=1+y{\log }_e\left(\frac{1+e^x}{2}\right)$
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   Answer: D

3. Q72020 · 2 Sept 2020 · Shift 1

   Let $y=y(x)$ be the solution of the differential equation, $\frac{2+\sin x}{y+1}\cdot \frac{dy}{dx}=-\cos x$, $y>0,y(0)=1$. If $y(\pi )=a$ and $\frac{dy}{dx}$ at $x=\pi$ is b, then the ordered pair (a, b) is equal to :

   1. A.$(2,1)$
   2. B.$\left(2,\frac{3}{2}\right)$
   3. C.$(1,-1)$
   4. D.$(1,1)$
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   Answer: D

4. Q72020 · 8 Jan 2020 · Shift 1

   Let $y=y(x)$ be a solution of the differential equation, $$\sqrt{1-x^2}\frac{dy}{dx}+\sqrt{1-y^2}=0,|x|<1.$$ If $y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$, then $y\left(-\frac{1}{\sqrt{2}}\right)$ is equal to :

   1. A.$-\frac{\sqrt{3}}{2}$
   2. B.None of those
   3. C.$\frac{1}{\sqrt{2}}$
   4. D.$-\frac{1}{\sqrt{2}}$
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   Answer: C

5. Q202020 · 8 Jan 2020 · Shift 2

   The differential equation of the family of curves, $x^2=4b(y+b),b\in \mathbf{R}$, is :

   1. A.$x(y^{\prime }{)}^2=x-2y{y}^{\prime }$
   2. B.$x(y^{\prime }{)}^2=2y{y}^{\prime }-x$
   3. C.$x{y}^{\prime \prime }=y^{\prime }$
   4. D.$x(y^{\prime }{)}^2=x+2y{y}^{\prime }$
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   Answer: D

6. Q152020 · 7 Jan 2020 · Shift 1

   If $y=y(x)$ is the solution of the differential equation, $e^y\left(\frac{dy}{dx}-1\right)=e^x$ such that $y(0)=0$, then $y(1)$ is equal to:

   1. A.$2+{\log }_{e}2$
   2. B.${\log }_{e}2$
   3. C.$1+{\log }_{e}2$
   4. D.$2e$
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   Answer: C

7. Q82020 · 7 Jan 2020 · Shift 2

   Let $y=y(x)$ be the solution curve of the differential equation, $(y^2-x)\frac{dy}{dx}=1$, satisfying $y(0)=1$. This curve intersects the x-axis at a point whose abscissa is :

   1. A.$2+e$
   2. B.$-e$
   3. C.2
   4. D.$2-e$
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   Answer: D