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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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781 Medium
781 of 975
Application +4 | -1
The solution of the differential equation 9ydydx+4x=09y\frac{{dy}}{{dx}} + 4x = 0 is
View answer and explanation
Correct option: B
y24+x29=c\frac{{{y^2}}}{4} + \frac{{{x^2}}}{9} = c
Option B (y24+x29=c\frac{{{y^2}}}{4} + \frac{{{x^2}}}{9} = c) is correct because it is the solution obtained by integrating the given differential equation.
782 Medium
782 of 975
Application +4 | -1
The solution of the differential equation dydx=x2+y2+12xy\frac{d y}{d x}=\frac{x^2+y^2+1}{2 x y} satisfying y(1)=1y(1)=1, is
View answer and explanation
Correct option: A
A hyperbola
The correct answer is A (A hyperbola) because the solution to the differential equation is a hyperbola.
783 Easy
783 of 975
Application +4 | -1
Differential equation whose solution is y=cx+c−c3y=c x+c-c^3, is
View answer and explanation
Correct option: B
y=xdydx+dydx−(dydx)3y=x \frac{d y}{d x}+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^3
Option B (y=xdydx+dydx−(dydx)3y=x \frac{{d y}}{{d x}}+\frac{{d y}}{{d x}}-\left(\frac{{d y}}{{d x}}\right)^3) is correct because substituting dydx=c\frac{{dy}}{{dx}} = c into the original equation yields this result.
784 Easy
784 of 975
Application +4 | -1
The solution of dydx=1+y+y2+x+xy+xy2\frac{{dy}}{{dx}} = 1 + y + {y^2} + x + xy + x{y^2} is
View answer and explanation
Correct option: D
4tan⁡−1(2y+13)=3(2x+x2)+c4{\tan ^{ - 1}}\left( {\frac{{2y + 1}}{{\sqrt 3 }}} \right) = \sqrt 3 \left( {2x + {x^2}} \right) + c
Option D (4tan⁡−1(2y+13)=3(2x+x2)+c4{\tan ^{ - 1}}\left( {\frac{{2y + 1}}{{\sqrt 3 }}} \right) = \sqrt 3 \left( {2x + {x^2}} \right) + c) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
785 Medium
785 of 975
Application +4 | -1
If ϕ(x)\phi(x) is a differentiable function then the solution of dy+(yϕ′(x)−ϕ(x)ϕ′(x))dx=0\mathrm{d} y+\left(y \phi^{\prime}(x)-\phi(x) \phi^{\prime}(x)\right) \mathrm{d} x=0 is
View answer and explanation
Correct option: A
y=(ϕ(x)−1)+Ce−ϕ(x)y=(\phi(x)-1)+C e^{-\phi(x)}
Option A (y=(ϕ(x)−1)+Ce−ϕ(x)y=(\phi(x)-1)+C e^{-\phi(x)}) is correct because it is the solution obtained by solving the given first-order linear differential equation using an integrating factor.
786 Easy
786 of 975
Application +4 | -1
The differential equation obtained by eliminating arbitrary constants from y=aebxy = a{e^{bx}} is
View answer and explanation
Correct option: C
yd2ydx2−(dydx)2=0y\frac{{{d^2}y}}{{d{x^2}}} - {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0
Option C (yd2ydx2−(dydx)2=0y\frac{{{d^2}y}}{{d{x^2}}} - {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0) is correct because it represents the differential equation after eliminating the arbitrary constants from the given equation.
787 Easy
787 of 975
Application +4 | -1
Differential equation of all circles with centres on y-axis is
View answer and explanation
Correct option: B
y1+y13−xy2=0y_{1}+y_{1}^{3}-x y_{2}=0
Option B (y1+y13−xy2=0y_{1}+y_{1}^{3}-x y_{2}=0) is correct because it represents the differential equation of all circles with centers on the y-axis.
788 Easy
788 of 975
Application +4 | -1
The solution of the differential equation ydydx=x−1y\frac{{dy}}{{dx}} = x - 1 satisfying y(1)=1y\left( 1 \right) = 1 is
View answer and explanation
Correct option: A
y2=x2−2x+2{y^2} = {x^2} - 2x + 2
Option A (y2=x2−2x+2{y^2} = {x^2} - 2x + 2) is correct because it is the solution to the given differential equation that satisfies the initial condition y(1) = 1.
789 Easy
789 of 975
Application +4 | -1
Let the solution curve of the differential equation xdydx−y=y2+16x2,y(1)=3x \frac{d y}{d x}-y=\sqrt{y^2+16 x^2}, y(1)=3 be y=y(x)y=y(x). Then y(2)y(2) is equal to :
View answer and explanation
Correct option: A
15
Option A (15) is correct because it is found by solving the given differential equation with the initial condition and then evaluating the solution at x=2.
790 Easy
790 of 975
Application +4 | -1
The integrating factor of the differential equation (ylog⁡y)dx=(log⁡y−x)dy\left( {y\log y} \right)dx = \left( {\log y - x} \right)dy is
View answer and explanation
Correct option: D
log⁡y\log y
Option D (log⁡y\log y) is correct because it is the integrating factor of the given differential equation after rewriting it in the standard form.
791 Medium
791 of 975
Application +4 | -1
The solution of differential equation y(2x4+y)dy+(4xy2−1)x2dx=0y\left(2 x^4+y\right) d y+\left(4 x y^2-1\right) x^2 d x=0 (where C{C} is an arbitrary constant)
View answer and explanation
Correct option: B
3x4y2+y3−x3=C3 x^4 y^2+y^3-x^3=C
Option B (3x4y2+y3−x3=C3 x^4 y^2+y^3-x^3=C) is correct because it is the solution obtained after rearranging and integrating the given differential equation.
792 Easy
792 of 975
Application +4 | -1
Let y=y(x)y = y\left( x \right) be the solution of the differential equation, 2+sin⁡xy+1.dydx=−cos⁡x,y>0, y(0)=1\frac{{2 + \sin x}}{{y + 1}}.\frac{{dy}}{{dx}} = - \cos x,y > 0,\,y\left( 0 \right) = 1. If y(π)=ay\left( \pi \right) = a and dydx\frac{{dy}}{{dx}} at x=πx = \pi is bb, then the ordered pair (a,b)\left( {a,b} \right) is equal to
View answer and explanation
Correct option: A
(1,1)\left( {1,1} \right)
The correct answer is A ((1,1)) because it is found by solving the differential equation and substituting the given conditions to find a and b.
793 Medium
793 of 975
Application +4 | -1
The solution of the differential equation

x2dydx⋅cos⁡1x−ysin⁡1x=−1,x^2 \frac{d y}{d x} \cdot \cos \frac{1}{x}-y \sin \frac{1}{x}=-1,

where y→−1y \rightarrow-1 as x→∞x \rightarrow \infty is
View answer and explanation
Correct option: A
y=sin⁡1x−cos⁡1xy=\sin \frac{1}{x}-\cos \frac{1}{x}
Option A (y=sin⁡1x−cos⁡1xy=\sin \frac{1}{x}-\cos \frac{1}{x}) is the correct solution obtained by solving the given differential equation using an integrating factor and applying the given condition.
794 Easy
794 of 975
Application +4 | -1
y=f(x){y}={f}({x}) passes through (1,2)(1,2) and satisfies the relation y(1+xy)dx−xdy=0{y}(1+x y) d x-x d y=0. Find value of 7f(12)7 f\left(\frac{1}{2}\right).
View answer and explanation
Correct option: A
4
Option A (4) is correct because solving the differential equation and using the given point (1,2), we find that 7f(1/2) = 4.
795 Medium
795 of 975
Application +4 | -1
Solution of the differential equation (x−y)2dy dx=(x-y)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}= a2a^2 is
View answer and explanation
Correct option: A
y=a2log⁡∣x−y−ax−y+a∣+cy=\frac{a}{2} \log \left|\frac{x-y-a}{x-y+a}\right|+c
Option A, y=a2log⁡∣x−y−ax−y+a∣+cy=\frac{a}{2} \log \left|\frac{x-y-a}{x-y+a}\right|+c, is the correct solution obtained by substituting x−y=ux-y=u and integrating the resulting equation.
796 Easy
796 of 975
Application +4 | -1
The solution of differential equation dydx=y2y(sin⁡y2+cos⁡y2)−x\frac{d y}{d x}=\frac{y}{2 y\left(\sin y^2+\cos y^2\right)-x} is
View answer and explanation
Correct option: A
xy+cos⁡y2−sin⁡y2=Cx y+\cos y^2-\sin y^2=C
Option A (xy+cos⁡y2−sin⁡y2=Cx y+\cos y^2-\sin y^2=C) is correct because it is the solution obtained after integrating the given differential equation using an integrating factor.
797 Hard
797 of 975
Application +4 | -1
If y(x)y(x) is the solution of the differential equation (xlog⁡x)dydx+y=2xlog⁡x(x≥1)(x \log x) \frac{d y}{d x}+y=2 x \log x(x \geq 1), then y(e)=……y(e)=\ldots \ldots
View answer and explanation
Correct option: C
2
Option C (2) is the correct value of y(e), obtained by solving the differential equation and applying the given condition.
798 Easy
798 of 975
Application +4 | -1
Differential equation whose sol. is y=cx+c−c3y=c x+c-c^3, is
View answer and explanation
Correct option: B
y=xdydx+dydx−(dydx)3y=x \frac{d y}{d x}+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^3
Option B (y=xdydx+dydx−(dydx)3y=x \frac{d y}{d x}+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^3) is correct because it is obtained by substituting c = dy/dx into the original equation.
799 Easy
799 of 975
Application +4 | -1
If y=f(x)y = f\left( x \right) is the equation of the curve an its differential equation is given by dydx=x+2y+3,\frac{{dy}}{{dx}} = \frac{{x + 2}}{{y + 3}}, then the equation of the curve, if it passes through (2,2), is
View answer and explanation
Correct option: A
x2−y2+4x−6y+4=0{x^2} - {y^2} + 4x - 6y + 4 = 0
Option A (x2−y2+4x−6y+4=0{x^2} - {y^2} + 4x - 6y + 4 = 0) is correct because it satisfies the given differential equation and passes through the point (2,2).
800 Easy
800 of 975
Application +4 | -1
General solution of the equation (1+x2)dydx+y=eTan⁡−1x\left(1+x^2\right) \frac{d y}{d x}+y=e^{\operatorname{Tan}^{-1} x} is
View answer and explanation
Correct option: C
2y=eTan⁡−1x+ce−Tan⁡−1x2 y=e^{\operatorname{Tan}^{-1} x}+c e^{-\operatorname{Tan}^{-1}} x
Option A (-12\frac{{1}}{{2}}) is correct because it is the value of y(1) obtained after solving the given differential equation with the initial condition y(0) = 1.
801 Medium
801 of 975
Application +4 | -1
The equation to the orthogonal trajectories of the system of parabolas y=ax2y=a x^2 is
View answer and explanation
Correct option: A
x22+y2=C\frac{x^2}{2}+y^2=C
Option A is correct because the orthogonal trajectories are found by replacing dy/dx with -dx/dy and solving the resulting differential equation.
802 Medium
802 of 975
Application +4 | -1
The solution of the differential equation x3dydx+4x2tan⁡y=exsec⁡yx^3 \frac{d y}{d x}+4 x^2 \tan y=e^x \sec y satisfying y(1)=0y(1)=0, is
View answer and explanation
Correct option: B
sin⁡y=ex(x−1)x−4\sin y=e^x(x-1) x^{-4}
Option D (sin⁡y=ex(x−1)x−4\sin y=e^x(x-1) x^{-4}) is the correct solution obtained by solving the differential equation with the given initial condition.
803 Medium
803 of 975
Application +4 | -1
The differential equation obtained on eliminating AA and BB from the equation y=Acos⁡ωt+Bsin⁡ωty = A\cos \omega t + B\sin \omega t is
View answer and explanation
Correct option: A
y2=−ω2y{y_2} = - {\omega ^2}y
Option A (y2=−ω2yy_2 = - {\omega ^2}y) is correct because it is the differential equation obtained by eliminating A and B from the given equation.
804 Easy
804 of 975
Application +4 | -1
Solution of the differential equation, ydx−xdy+xy2dx=0y d x-x d y+x y^2 d x=0 can be
View answer and explanation
Correct option: A
2x+x2y=λy2 x+x^2 y=\lambda y
Option A (2x+x2y=λy2 x+x^2 y=\lambda y) is the correct solution, obtained by rearranging and integrating the given differential equation.
805 Easy
805 of 975
Understanding +4 | -1
For the differential equation d2ydx2+y=0\frac{d^2 y}{d x^2}+y=0, let its solution be
y=ϕ1(x)=2sin⁡(x+π4)y=\phi_1(x)=2 \sin \left(x+\frac{\pi}{4}\right).

Assertion: The function y=ϕ1(x)\mathrm{y}=\phi_1(\mathrm{x}) is called the particular solution.

Reason: The solution which is free from arbitrary constant, is called a particular solution.
View answer and explanation
Correct option: A
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because a particular solution is free from arbitrary constants.
806 Medium
806 of 975
Application +4 | -1
If y=f(x)y=f(x) satisfies the differential equation sin⁡xdydx+2ycos⁡x=8\sin x \frac{d y}{d x}+2 y \cos x=8 with f(π/2)=8f(\pi / 2)=8, then the minimum value of f(x)f(x) is:
View answer and explanation
Correct option: A
4
Option A (4) is correct because the minimum value of the function f(x)f(x) is found by solving the differential equation and analyzing the resulting function.
807 Medium
807 of 975
Application +4 | -1
The integrating factor of the differential equation dydx+1x.y=3x\frac{{dy}}{{dx}} + \frac{1}{x}.y = 3x is
View answer and explanation
Correct option: A
xx
Option A (xx) is correct because it is the integrating factor calculated using the formula e∫Pdxe^{\int P dx} for the given differential equation.
808 Medium
808 of 975
Application +4 | -1
A particular solution of the initial value differential equation log⁡dydx=7x+8y,y(0)=0\log \frac{d y}{d x}=7 x+8 y, y(0)=0 is
View answer and explanation
Correct option: D
7e−8y+8e7x=157 e^{-8 y}+8 e^{7 x}=15
The correct answer is C (8e−8y+7e7x=158 e^{-8 y}+7 e^{7 x}=15) because it satisfies the given differential equation and the initial condition y(0)=0.
809 Medium
809 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation sec⁡xdy+{2(1−x)tan⁡x+x(2−x)}dx=0\sec x d y+\{2(1-x) \tan x+x(2-x)\} d x=0 such that y(0)=2y(0)=2. Then y(2)y(2) is equal to :
View answer and explanation
Correct option: D
2
The correct answer is D (2) because solving the differential equation and applying the initial condition leads to y(2) = 2.
810 Easy
810 of 975
Application +4 | -1
Consider the differential equation, y2dx+(x−1y)dy=0.{y^2}dx + \left( {x - \frac{1}{y}} \right)dy = 0. if value of yy is 1 when x=1, then the value of xx for which y=2y = 2 is
View answer and explanation
Correct option: B
32−1e\frac{3}{2} - \frac{1}{{\sqrt e }}
Option B (32−1e\frac{3}{2} - \frac{1}{{\sqrt e }}) is correct because it is the value of x when y=2 after solving the given differential equation.
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