Differential Equation – JEE Main Previous Year Question with Solutions
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Q. The differential equation which represents the family of curves $y=c_{1} e^{c_{2} x},$ where $c_{1}$ and $c_{2}$ are arbitrary constants, is :-(1) yy” = y’(2) $\mathrm{yy}^{\prime \prime}=\left(\mathrm{y}^{\prime}\right)^{2}$(3) $y^{\prime}=y^{2}$(4) $y^{\prime \prime}=y^{\prime} y$ [AIEEE-2009]

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Sol. (2)

Q. Solution of the differential equation $\cos x$ dy $=y(\sin x-y) d x, 0<x<\frac{\pi}{2}$ is :(1) sec x = (tan x + c) y(2) y sec x = tan x + c(3) y tan x = sec x + c(4) tan x = (sec x + c) y [AIEEE-2010]

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Sol. (1)

Q. If $\frac{d y}{d x}=y+3>0$ and $y(0)=2,$ then $y(\ln 2)$ is equal to $:-$(1) 13              (2) –2              (3) 7               (4) 5 [AIEEE-2011]

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Sol. (3)$\frac{d y}{d x}=y+3>0 \quad y(0)=2, y(\log 2)=?$$\int \frac{d y}{y+3}=\int d x$$\log |y+3|=x+c$$\mathrm{y}(0)=2$$\log |2+3|=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5$y. $(\log 2)=?$$\log |y+3|=\log 2+\log 5$$\log |y+3|=\log 10$$y+3=10$$y=7$

Q. Let I be the purchase value of an equipment and $V(t)$ be the value after it has been used for t years. The value $\mathrm{V}(\mathrm{t})$ depreciates at a rate given by differential equation $\frac{\mathrm{dV}(\mathrm{t})}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{t})$ where $\mathrm{k}>0$ is a constant and $\mathrm{T}$ is the total life in years of the equipment. Then the scrap value $\mathrm{V}(\mathrm{T})$ of the equipment is :-.(1) $\mathrm{I}-\frac{\mathrm{k}(\mathrm{T}-\mathrm{t})^{2}}{2}$(2) $\mathrm{e}^{-\mathrm{kT}}$(3) $\mathrm{T}^{2}-\frac{\mathrm{I}}{\mathrm{k}}$(4) $\mathrm{I}-\frac{\mathrm{k} \mathrm{T}^{2}}{2}$ [AIEEE-2011]

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Sol. (4)$\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{t})$$\int \mathrm{d} \mathrm{V}=\int-\mathrm{k}(\mathrm{T}-\mathrm{t}) \mathrm{dt}$$V=-k\left[T t-\frac{t^{2}}{2}\right]+C$At $\mathrm{t}=0 \quad \mathrm{V}=\mathrm{I} \Rightarrow \mathrm{C}=\mathrm{I}$$\mathrm{V}=-\mathrm{kt}\left(\mathrm{T}-\frac{\mathrm{t}}{2}\right)+\mathrm{I}$$V(T)=-k T\left(T-\frac{T}{2}\right)+I=\frac{-k T^{2}}{2}+I$

Q. The curve that passes through the point (2, 3), and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by :(1) $\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{3}\right)^{2}=2$(2) 2y – 3x = 0(3) $y=\frac{6}{x}$(4) $x^{2}+y^{2}=13$ [AIEEE-2011]

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Sol. (3)Equation of tangent at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is$y-y_{1}=\frac{d y_{1}}{d x_{1}}\left(x-x_{1}\right)$$\mathrm{x}- intercept =\mathrm{x}_{1}-\mathrm{y}_{1} \frac{\mathrm{d} \mathrm{x}_{1}}{\mathrm{d} y_{1}}According to questionx_{1}=\frac{x_{1}-y_{1} \frac{d x_{1}}{d y_{1}}}{2}$$\Rightarrow \mathrm{x}_{1}=-\mathrm{y}_{1} \frac{\mathrm{d} \mathrm{x}_{1}}{\mathrm{d} y_{1}}$$\int \frac{d y}{y}=\int-\frac{d x}{x}$$\Rightarrow \quad \ell n y=-\ell n x+\ell n c$$\Rightarrow \quad \mathrm{y}=\frac{\mathrm{c}}{\mathrm{x}} \quad \Rightarrow \quad \mathrm{xy}=\mathrm{c}Now at x=2, y=3$$\Rightarrow \quad c=6$$\therefore \quad x y=6 \quad \Rightarrow \quad y=\frac{6}{x} Q. Consider the differential equation \mathrm{y}^{2} \mathrm{d} \mathrm{x}+\left(\mathrm{x}-\frac{1}{\mathrm{y}}\right) \mathrm{d} \mathrm{y}=0 . It \mathrm{y}(1)=1, then \mathrm{x} is given by :(1) 1-\frac{1}{y}+\frac{\frac{1}{e^{y}}}{e}(2) 4-\frac{2}{y}-\frac{\frac{1}{e^{y}}}{e}(3) 3-\frac{1}{y}+\frac{\frac{1}{e^{y}}}{e}(4) 1+\frac{1}{y}-\frac{\frac{1}{e^{y}}}{e} [AIEEE-2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\mathrm{y}^{2} \mathrm{d} \mathrm{x}+\left(\mathrm{x}-\frac{1}{\mathrm{y}}\right) \mathrm{d} \mathrm{y}=0$$\Rightarrow \quad \mathrm{y}^{2} \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} y}+\mathrm{x}=\frac{1}{\mathrm{y}} \quad \Rightarrow \quad \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} y}+\frac{\mathrm{x}}{\mathrm{y}^{2}}=\frac{1}{\mathrm{y}^{3}}$Integrating factor (I.F.) $=\mathrm{e}^{\int \frac{1}{\mathrm{y}^{2}} \mathrm{d} \mathrm{y}}=\mathrm{e}^{-1 / \mathrm{y}}$General solution is –$\mathbf{x} \cdot \mathrm{e}^{-1 / \mathrm{y}}=\int \frac{1}{y^{3}} e^{-1 / y} \mathrm{d} y+c$Let $\mathrm{I}_{1}=\int \frac{1}{y^{3}} e^{-1 / y} \mathrm{d} y$put $\quad \frac{-1}{y}=t$$y^{-2} d y=d t$$\therefore \quad \mathrm{I}_{1}=-\int \mathrm{te}^{\mathrm{t}} \mathrm{dt}$$=-e^{t}(t-1)$$=e^{t}(1-t)$General solution is$\mathrm{xe}^{-1 / \mathrm{y}}=\mathrm{e}^{-1 / \mathrm{y}}\left(1+\frac{1}{\mathrm{y}}\right)+\mathrm{C}$$\Rightarrow \quad \mathrm{x}=1+\frac{1}{y}+\mathrm{Ce}^{1 / \mathrm{y}}Put x = 1, y = 1\therefore \quad 1=1+\frac{1}{1}+\mathrm{Ce}^{1 / 1}$$\Rightarrow \quad C=-1 / \mathrm{e}$$\therefore \quad x=1+\frac{1}{y}-\frac{e^{1 / y}}{e} Q. The population p(t) at time t of a certain mouse species satisfies the differential equation \mathrm{dp}(\mathrm{t}) \mathrm{dt} = 0.5 p(t) – 450. If p(0) = 850, then the time at which the population becomes zero is :(1) ln18 (2) 2 ln18 (3) ln9 (4) \frac{1}{2} \ln 18 [AIEEE-2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=\frac{1}{2} \mathrm{p}(\mathrm{t})-450integrate\int \frac{\mathrm{d} \mathrm{p}}{\mathrm{p}-900}=\int \frac{1}{2} \mathrm{d} \mathrm{t}$$\ell \mathrm{n}|(\mathrm{p}-900)|=\frac{1}{2} \mathrm{t}+\mathrm{C} \quad \ldots .(1)$given $\mathrm{t}=0 \rightarrow \mathrm{p}=850 \quad \therefore \mathrm{C}=\ell \mathrm{n} 50$from ( 1)

Q. At present a firm is manufacturing 2000 items. It is estimated that the rate of change ofproduction P w.r.t. additional number of workers x is given by $\frac{d P}{d x}=100-12 \sqrt{x}$. If the firm employs 25 more workers, then the new level of production of items is :(1) 2500              (2) 3000               (3) 3500                  (4) 4500 [JEE (Main)-2013]

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Sol. (3)$\mathrm{P}=100 \mathrm{x}-12 \mathrm{x}^{3 / 2} \cdot \frac{2}{3}+\mathrm{C}$x = 0,P = 2000C = 2000$\mathrm{P}_{(\mathrm{x}=25)}=2500-1000+2000=3500$

Q. If the surface area of a sphere of radius r is increasing uniformly at the rate $8 \mathrm{cm}^{2} / \mathrm{s}$, then the rate of change of its volume is :(1) proportional to $r^{2}$(2) constant(3) proportional to r(4) proportional to $\sqrt{\mathrm{r}}$ [JEE-Main (On line)-2013]

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Sol. (3)$\mathrm{S}=4 \pi \mathrm{r}^{2}$$\frac{\mathrm{ds}}{\mathrm{dt}}=8=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \& \quad \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^{3}$$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}$$\Rightarrow \frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=4 \mathrm{r} Q. Consider the differential equation \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{\mathrm{y}^{3}}{2\left(\mathrm{xy}^{2}-\mathrm{x}^{2}\right)}Statement 1 : The substitution z =y^{2} transforms the above equation into a first order homogenous differential equation.Statement 2 : The solution of this differential equation is y^{2} e^{-\frac{y^{2}}{x}}=C.(1) Statement 1 is false and statement 2 is true. (2) Both statements are true.(3) Statement 1 is true and statement 2 is false. (4) Both statements are false. [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)put z=y^{2}$$\Rightarrow \quad \frac{\mathrm{d} z}{\mathrm{dx}}=2 \mathrm{y} \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}$$\Rightarrow \quad D.E. is \frac{\mathrm{d} z}{\mathrm{dx}}=\frac{2 \mathrm{z}^{2}}{2\left(\mathrm{xz}-\mathrm{x}^{2}\right)}$$\Rightarrow \quad z(x d z-z d x)=x^{2} d z$$\Rightarrow \quad \int d\left(\frac{z}{x}\right)=\int \frac{d z}{z}$$\Rightarrow \quad \frac{\mathrm{z}}{\mathrm{x}}=\ell \mathrm{nz}+\ell \mathrm{nC}$

Q. If a curve passes through the point $\left(2, \frac{7}{2}\right)$ and has slope $\left(1-\frac{1}{\mathrm{x}^{2}}\right)$ at any point $(\mathrm{x}, \mathrm{y})$ on it, then the ordinate of the point on the curve whose abscissa is $-2$ is :$(1)-\frac{5}{2}$( 2)$\frac{5}{2}$$(3)-\frac{3}{2}( 4)\frac{3}{2} [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)\frac{\mathrm{d} y}{\mathrm{dx}}=1-\frac{1}{\mathrm{x}^{2}}$$\Rightarrow \quad \mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+1$$\Rightarrow \quad \mathrm{y}=-2-\frac{1}{2}+1=-\frac{3}{2} Q. The equation of the curve passing through the origin and satisfying the differential equation \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2} is :(1)\left(1+x^{2}\right) y=x^{3}(2) 3\left(1+x^{2}\right) y=4 x^{3}(3) 3\left(1+x^{2}\right) y=2 x^{3}(4) \left(1+x^{2}\right) y=3 x^{3} [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)\int \mathrm{d}\left(\left(1+\mathrm{x}^{2}\right) \mathrm{y}\right)=\int 4 \mathrm{x}^{2} \mathrm{d} \mathrm{x}$$\Rightarrow \quad \mathrm{y}\left(1+\mathrm{x}^{2}\right)=\frac{4 \mathrm{x}^{3}}{3}+\mathrm{C}$ passes through $(0,0)$$\Rightarrow \quad 3 \mathrm{y}\left(1+\mathrm{x}^{2}\right)=4 \mathrm{x}^{3} Q. Let the population of rabbits surviving at a time t be governed by the differential equation \frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=\frac{1}{2} \mathrm{p}(\mathrm{t})-200 . If \mathrm{p}(0)=100, then \mathrm{p}(\mathrm{t}) equals :(1) 400-300 \mathrm{e}^{\mathrm{t} / 2}(2) 300-200 \mathrm{e}^{-t / 2}(3) 600-500 \mathrm{e}^{t / 2}(4) 400-300 \mathrm{e}^{-1 / 2} [JEE(Main)-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=\frac{1}{2} \mathrm{p}(\mathrm{t})-200$$\int_{10}^{p(t)} \frac{d p(t)}{p(t)-400}=\int_{0}^{t} \frac{d t}{2} \Rightarrow \log \left|\frac{p(t)-400}{-300}\right|=\frac{t}{2}$$|p(t)-400|=300 e^{t / 2}$$400-p(t)=300 e^{t / 2}$$\therefore \mathrm{p}(\mathrm{t})=400-300 \mathrm{e}^{\mathrm{t} / 2} Q. Let \mathrm{y}(\mathrm{x}) be the solution of the differential equation (\mathrm{x} \log \mathrm{x}) \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}+\mathrm{y}=2 \mathrm{x} \log \mathrm{x},(\mathrm{x} \geq 1) Then \mathrm{y}(\mathrm{e}) is equal to :(1) 2 (2) 2e (3) e (4) 0 [JEE(Main)-2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)(x \log x) \frac{d y}{d x}+y=2 x \log x,(x \geq 1)$$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=2$I.F. $=e^{\int \frac{d x}{x \log x}}=e^{\log (\log x)}=\log x$Solution is $y \log x=\int 2 \log x d x$

Q. If a curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ passes through the point $(1,-1)$ and satisfies the differential equation, $\mathrm{y}(1+\mathrm{xy}) \mathrm{dx}=\mathrm{x}$ dy, then $\mathrm{f}\left(-\frac{1}{2}\right)$ is equal to :( 1)$\frac{4}{5}$(2) $-\frac{2}{5}$(3) $-\frac{4}{5}$( 4)$\frac{2}{5}$ [JEE(Main)-2016]

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Sol. (1)Given differential equation$\mathrm{ydx}+\mathrm{xy}^{2} \mathrm{dx}=\mathrm{xdy}$$\Rightarrow \frac{\mathrm{xdy}-\mathrm{ydx}}{\mathrm{y}^{2}}=\mathrm{xdx}$$\Rightarrow-\mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{d}\left(\frac{\mathrm{x}^{2}}{2}\right)$Integrating we get$-\frac{x}{y}=\frac{x^{2}}{2}+C$$\because It passes through (1,-1)$$\therefore \quad 1=\frac{1}{2}+\mathrm{C} \Rightarrow \mathrm{C}=\frac{1}{2}$$\therefore \quad x^{2}+1+\frac{2 x}{y}=0 \Rightarrow y=\frac{-2 x}{x^{2}+1}$$\therefore \mathrm{f}\left(-\frac{1}{2}\right)=\frac{4}{5}$

Q. If $(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$ and $y(0)=1,$ then $y\left(\frac{\pi}{2}\right)$ is equal to :-(1) $\frac{4}{3}$( 2)$\frac{1}{3}$(3) $-\frac{2}{3}$$(4)-\frac{1}{3} [JEE(Main)-2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$$\frac{d}{d x}(2+\sin x)(y+1)=0$$(2+\sin x)(y+1)=c$$x=0, y=1 \Rightarrow c=4$$\mathrm{y}+1=\frac{4}{2+\sin \mathrm{x}}$$y\left(\frac{\pi}{2}\right)=\frac{4}{3}-1=\frac{1}{3}$

Q. Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\sin \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \cos \mathrm{x}=4 \mathrm{x}, \mathrm{x} \in(0$ \pi).If $\mathrm{y}\left(\frac{\pi}{2}\right)=0,$ then $\mathrm{y}\left(\frac{\pi}{6}\right)$ is equal to :(1) $\frac{-8}{9 \sqrt{3}} \pi^{2}$(2) $-\frac{8}{9} \pi^{2}$(3) $-\frac{4}{9} \pi^{2}$(4) $\frac{4}{9 \sqrt{3}} \pi^{2}$ [JEE(Main)-2018]

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Sol. (2)

• April 25, 2021 at 1:39 am

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• February 22, 2021 at 4:27 pm

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• February 22, 2021 at 4:28 pm

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• December 3, 2020 at 5:50 pm

should be more questions included

• August 10, 2021 at 8:00 am

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• December 2, 2020 at 12:29 pm

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• October 29, 2020 at 11:41 pm

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• October 19, 2020 at 7:14 pm

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• October 1, 2020 at 11:34 pm

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• December 24, 2020 at 8:41 pm

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• September 23, 2020 at 2:48 pm

please give 2019 and 2020 questions also

• August 20, 2020 at 10:52 pm

Thank you so much

• November 2, 2020 at 3:05 pm

No solution available due to the error

• November 5, 2020 at 6:33 pm

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• November 6, 2020 at 10:14 am