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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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61 Easy
61 of 975
Application +4 | -1
The differential equation of family of the curves x2+y2=2axx^2+y^2=2 a x is ..... . ( aa is an arbitrary constant)
View answer and explanation
Correct option: B
2xydydx+x2−y2=02 x y \frac{d y}{d x}+x^2 - y^2=0
Option B (2xydydx+x2−y2=02xy \frac{{dy}}{{dx}} + x^2 - y^2 = 0) is correct because it is the differential equation obtained by eliminating the arbitrary constant 'a' from the given family of curves.
62 Easy
62 of 975
Application +4 | -1
Solution of differential equation (x+3y2)dydx=y,(y>0)\left(x+3 y^2\right) \frac{d y}{d x}=y,(y>0) is
View answer and explanation
Correct option: A
x=3y2+cyx=3 y^2+c y
Option A, x=3y2+cyx=3 y^2+c y, is the correct solution obtained by rearranging the equation and solving the differential equation.
63 Easy
63 of 975
Application +4 | -1
If ϕ(x)=ϕ′(x)\phi \left( x \right) = \phi '\left( x \right) and ϕ(1)=2\phi \left( 1 \right) = 2, then ϕ(3)\phi \left( 3 \right) equals
View answer and explanation
Correct option: B
2e22{e^2}
Option B (2e22{e^2}) is correct because it is the value of ϕ(3)\phi(3) obtained by solving the differential equation and applying the given condition.
64 Medium
64 of 975
Application +4 | -1
The equation of curve passing through the point (1,π4)\left( {1,\frac{\pi }{4}} \right) and having slope of tangent at any point (x,y)\left( {x,y} \right) as yx−cos⁡2(yx)\frac{y}{x} - {\cos ^2}\left( {\frac{y}{x}} \right), is
View answer and explanation
Correct option: B
x=e1−tan⁡(yx)x = {e^{1 - \tan \left( {\frac{y}{x}} \right)}}
Option B (x=e1−tan⁡(yx)x = {e^{1 - \tan \left( {\frac{y}{x}} \right)}}) is correct because it is the solution of the given differential equation after substituting v=yxv = \frac{y}{x} and applying the initial condition.
65 Medium
65 of 975
Analyzing +4 | -1
The differential equation of the family y=aex+bxex+cx2exy = a{e^x} + bx{e^x} + c{x^2}{e^x} of curves, where a,b,ca,b,c are arbitrary constants, is
View answer and explanation
Correct option: D
y′′′−3y′+3y′−y=0y''' - 3y' + 3y' - y = 0
Option D (y′′′−3y′′+3y′−y=0y''' - 3y'' + 3y' - y = 0) is correct because it represents the differential equation obtained after differentiating the given equation thrice and eliminating the arbitrary constants a, b, and c.
66 Medium
66 of 975
Application +4 | -1
If  y=sin⁡(mSin⁡−1x),then   < br > (1−x2)d2ydx2={\text{If}}\,\,y = \sin (m{\operatorname{Sin} ^{ - 1}}x),{\text{then}}\,\,{\text{ < br > }}(1 - {x^2})\frac{{{d^2}y}}{{d{x^2}}} =
View answer and explanation
Correct option: B
xdydx−m2yx \frac{d y}{d x}-m^2 y
Option B is correct because the given equation satisfies the differential equation (1−x2)d2ydx2=xdydx−m2y(1 - {x^2})\frac{{{d^2}y}}{{d{x^2}}} = x \frac{d y}{d x}-m^2 y.
67 Medium
67 of 975
Application +4 | -1
The solution of differential equation 2x3ydy+(1−y2)(x2y2+y2−1)dx=02 x^3 y d y+\left(1-y^2\right)\left(x^2 y^2+y^2-1\right) d x=0 is:
View answer and explanation
Correct option: C
x2y2=(cx−1)(1−y2)x^2 y^2=(c x-1)\left(1-y^2\right)
Option C (x2y2=(cx−1)(1−y2)x^2 y^2=(c x-1)(1-y^2)) is correct because it is the solution obtained after solving the given differential equation using appropriate substitutions and integrating factor.
68 Medium
68 of 975
Understanding +4 | -1
If f(x)f(x) and g(x)g(x) are two solutions of the differential equation ad2ydx2+x2dydx+y=exa \frac{d^2 y}{d x^2}+x^2 \frac{d y}{d x}+y=e^x, then f(x)−g(x)f(x)-g(x) is the solution of
View answer and explanation
Correct option: D
ad2ydx2+x2dydx+y=0a \frac{d^2 y}{d x^2}+x^2 \frac{d y}{d x}+y=0
Option D (ad2ydx2+x2dydx+y=0a \frac{d^2 y}{d x^2}+x^2 \frac{d y}{d x}+y=0) is correct because the difference of two solutions to a non-homogeneous linear differential equation is a solution to the corresponding homogeneous equation.
69 Easy
69 of 975
Application +4 | -1
A curve that passes through (2,4)(2,4) and having subnormal of constant length of 8 units can be
View answer and explanation
Correct option: A
y2=16x−8y^2=16 x-8
Option A (y2=16x−8y^2=16 x-8) is the correct answer because it satisfies the given condition of having a constant subnormal length of 8 units and passes through the point (2,4).
70 Medium
70 of 975
Application +4 | -1
The solution of the differential equation dydx=1xy(x2sin⁡y2+1)\frac{d y}{d x}=\frac{1}{x y\left(x^2 \sin y^2+1\right)} is
View answer and explanation
Correct option: A
x2(cos⁡y2−sin⁡y2−2ce−y2)=2x^2\left(\cos y^2-\sin y^2-2 c e^{-y^2}\right)=2
Option A, x2(cos⁡y2−sin⁡y2−2ce−y2)=2x^2(\cos y^2-\sin y^2-2 c e^{-y^2})=2, is the correct solution obtained by solving the given differential equation using an appropriate substitution and integrating factor.
71 Easy
71 of 975
Application +4 | -1
If a curve passes through the point (2,72)\left(2, \frac{7}{2}\right) and has slope (1−1x2)\left(1-\frac{1}{x^2}\right) at any point (x,y)(x, y) on it, then the ordinate of the point on the curve whose abscissa is -2 is :
View answer and explanation
Correct option: A
−32-\frac{3}{2}
Option A (−32-\frac{3}{2}) is correct because it is the ordinate of the point on the curve whose abscissa is -2, obtained after solving the given differential equation and applying the initial condition.
72 Easy
72 of 975
Application +4 | -1
The solution of the differential equation dydx=ytan⁡x−2sin⁡x\frac{{dy}}{{dx}} = y\tan x - 2\sin x, is
View answer and explanation
Correct option: D
ycos⁡x=c+12cos⁡2xy\cos x = c + \frac{1}{2}\cos 2x
Option B (ycos⁡x=c+12sin⁡2xy\cos x = c + \frac{1}{2}\sin 2x) is correct because it is the solution obtained after finding the integrating factor and integrating the differential equation.
73 Medium
73 of 975
Application +4 | -1
A curve passes through the point (0,1)\left( {0,1} \right) and the gradient at (x,y)\left( {x,y} \right) on it is y(xy−1)y\left( {xy - 1} \right). The equation of the curve is
View answer and explanation
Correct option: B
y(x+1)=1y\left( {x + 1} \right) = 1
Option B (y(x+1)=1y(x + 1) = 1) is correct because it satisfies the given differential equation and the initial condition (0, 1).
74 Easy
74 of 975
Application +4 | -1
The equation of the curve through the point (3,2) and whose slope x2y+1\frac{{{x^2}}}{{y + 1}}, is
View answer and explanation
Correct option: D
y22+y=x33−5\frac{{{y^2}}}{2} + y = \frac{{{x^3}}}{3} - 5
Option D (y22+y=x33−5\frac{{{y^2}}}{2} + y = \frac{{{x^3}}}{3} - 5) is correct because it is the solution to the differential equation that passes through the point (3,2).
75 Medium
75 of 975
Application +4 | -1
The solution of (1+sin⁡2x)dy−cos⁡x(1+y2)dx=0\left(1+\sin ^2 x\right) d y-\cos x\left(1+y^2\right) d x=0 given that y=2y=2 when, x=π2x=\frac{\pi}{2}
View answer and explanation
Correct option: B
y=3sin⁡x+13−sin⁡xy=\frac{3 \sin x+1}{3-\sin x}
Option B (y=3sin⁡x+13−sin⁡xy=\frac{3 \sin x+1}{3-\sin x}) is correct because it is the solution to the given differential equation that satisfies the initial condition y=2y=2 when x=π2x=\frac{\pi}{2}.
76 Easy
76 of 975
Application +4 | -1
∫(x4−x)1/4x5dx\int \frac{\left(x^4-x\right)^{1 / 4}}{x^5} d x is equal to : (where c\mathrm{c} is constant of integration)
View answer and explanation
Correct option: A
415(1−1x3)5/4+c\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{5 / 4}+c
Option A (415(1−1x3)5/4+c\frac{{4}}{{15}}\left(1-\frac{{1}}{{x^3}}\right)^{{5 / 4}}+c) is correct because it is the result of the integration after a suitable substitution.
77 Medium
77 of 975
Application +4 | -1
The solution of the differential equation cos⁡ydx+(1+2e−x)sin⁡ydy=0\cos y d x+\left(1+2 e^{-x}\right) \sin y d y=0, where y(0)=π4y(0)=\frac{\pi}{4}, is
View answer and explanation
Correct option: A
ex+2=32cos⁡ye^x+2=3 \sqrt{2} \cos y
Option A (ex+2=32cos⁡ye^x+2=3 \sqrt{2} \cos y) is the correct solution after separating variables, integrating, and applying the initial condition.
78 Hard
78 of 975
Application +4 | -1
The solution of the differential equation (1+y2)+(x−etan⁡x−1y)dydx=0\left(1+y^2\right)+\left(x-e^{\operatorname{tan} x^{-1} y}\right) \frac{d y}{d x}=0 is ...... .
View answer and explanation
Correct option: A
2xetan⁡−1y=e2tan⁡−1y+c2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c
Option A (2xetan⁡−1y=e2tan⁡−1y+c2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c) is the correct solution, found by rearranging the equation and using an integrating factor.
79 Medium
79 of 975
Application +4 | -1
If xdydx=y(log⁡y−log⁡x+1)x \frac{d y}{d x}=y(\log y-\log x+1), then the solution of the equation is
View answer and explanation
Correct option: C
log⁡(yx)=cx\log \left(\frac{y}{x}\right)=c x
Option C (log⁡(yx)=cx\log (\frac{y}{x})=c x) is correct because it is the solution obtained after solving the given homogeneous differential equation using the substitution y=vxy=vx.
80 Easy
80 of 975
Application +4 | -1
Which of the following is true for y(x) that satisfies the differential equation dydx=xy−1+x−y;y(0)=0\frac{{dy}}{{dx}} = \mathrm{ }xy\mathrm{ }-\mathrm{ }1\mathrm{ } + \mathrm{ }x\mathrm{ }-\mathrm{ }y;\mathrm{ }y\mathrm{ }\left( 0 \right)\mathrm{ } = 0
View answer and explanation
Correct option: A
y(1)=e−12−1y\mathrm{ }\left( 1 \right)\mathrm{ } = {e^{ - \frac{1}{2}}} - 1
Option A (y(1)=e−12−1y\mathrm{ }\left( 1 \right)\mathrm{ } = {e^{ - \frac{1}{2}}} - 1) is correct because it is the solution to the differential equation with the given initial condition.
81 Hard
81 of 975
Application +4 | -1
The rate of decay of a radioactive body at any time is proportional to its mass present at that time. If the mass of body is 100gms100 \mathrm{gms} after one hour and 80gms80 \mathrm{gms} after two hours, then the initial mass of radioactive body is ........ .
View answer and explanation
Correct option: C
125
Option C (125) is correct because it is the initial mass calculated using the exponential decay model and the given data points.
82 Easy
82 of 975
Understanding +4 | -1
The order of the differential equation of the family of rectangular hyperbolas having fixed centre, is
View answer and explanation
Correct option: A
2
Option 2 is correct because the family of rectangular hyperbolas with a fixed center requires two parameters to define it.
83 Easy
83 of 975
Application +4 | -1
Solution of the differential equation x dyx2+y2=(yx2+y2−1)dx\frac{x \,d y}{x^2+y^2}=\left(\frac{y}{x^2+y^2}-1\right) d x, is
View answer and explanation
Correct option: A
tan⁡−1(yx)=−x+C\tan ^{-1}\left(\frac{y}{x}\right)=-x+C
Option A (tan⁡−1(yx)=−x+C\tan ^{-1}\left(\frac{y}{x}\right)=-x+C) is correct because it is the solution obtained by integrating the given differential equation after rearrangement.
84 Easy
84 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation cos⁡xdydx+ysin⁡x=1,x∈(0,π2)\cos x \frac{d y}{d x}+y \sin x=1, x \in\left(0, \frac{\pi}{2}\right). If y(π4)=0{y}\left(\frac{\pi}{4}\right)=0, then y(π6){y}\left(\frac{\pi}{6}\right) is
View answer and explanation
Correct option: C
1−32\frac{1-\sqrt{3}}{2}
Option C (1−32\frac{{1-\sqrt{{3}}}}{{2}}) is correct because it is the value of y(π6\frac{{\pi}}{{6}}) obtained after solving the given differential equation and applying the initial condition.
85 Easy
85 of 975
Application +4 | -1
The solution of dydx+y=e−x,y(0)=0\frac{{dy}}{{dx}} + y = {e^{ - x}},y\left( 0 \right) = 0 is
View answer and explanation
Correct option: D
y=xe−xy = x{e^{ - x}}
Option D (y=xe−xy = x{e^{ - x}}) is correct because it satisfies the given differential equation and the initial condition y(0) = 0.
86 Medium
86 of 975
Application +4 | -1
The solution of the differential equation dydx+yx=x2\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}, is
View answer and explanation
Correct option: C
y=x34+Cx−1y = \frac{{{x^3}}}{4} + C{x^{ - 1}}
Option C (y=x34+Cx−1y = \frac{{{x^3}}}{4} + C{x^{ - 1}}) is correct because it's the solution obtained by finding the integrating factor and solving the linear differential equation.
87 Medium
87 of 975
Analyzing +4 | -1
If y:[1,∞)→[1,∞)y:[1, \infty) \rightarrow[1, \infty) is a function of x\mathrm{x} such that 2x=y1/5+y−1/52 x=y^{1 / 5}+y^{-1 / 5} and (x2−1)d2ydx2+xdydx=ky(k∈R)\left(x^2-1\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=k y(k \in R) Then k=\mathrm{k}=
View answer and explanation
Correct option: C
25
The correct answer is 25 because after differentiating the given equation and simplifying, the value of k in the differential equation is found to be 25.
88 Medium
88 of 975
Application +4 | -1
Solution of the differential equation dydx=y3+y2y2−x2x3\frac{d y}{d x}=\frac{y^3+y^2 \sqrt{y^2-x^2}}{x^3} is
View answer and explanation
Correct option: A
y+y2−x2=kxyy+\sqrt{y^2-x^2}=k x y
Option A (y+y2−x2=kxyy+\sqrt{y^2-x^2}=k x y) is correct because it is the solution obtained after solving the homogeneous differential equation using the substitution y=vxy=vx.
89 Easy
89 of 975
Application +4 | -1
Suppose yy is a function of xx that satisfies dydx=1−y2x2\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{x^2} and y=0y=0 at x=2πx=\frac{2}{\pi}, then y(3π)y\left(\frac{3}{\pi}\right) is equal to:
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because it's found by solving the differential equation and using the given initial condition to find the constant of integration.
90 Medium
90 of 975
Application +4 | -1
The solution of the differential equation dydx+ycot⁡x=2cos⁡x\frac{d y}{d x}+y \cot x=2 \cos x is
View answer and explanation
Correct option: D
2ysin⁡x+cos⁡2x=c2 y \sin x+\cos 2 x=c
Option D (2ysin⁡x+cos⁡2x=c2 y \sin x+\cos 2 x=c) is correct because it represents the solution to the given linear differential equation after finding the integrating factor and integrating.
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