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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Medium

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Practice format and chapter quiz

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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Showing 91-120 of 452 questions

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91 Medium
91 of 452
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 - 2ce^{-y^2}) = 2) is correct because it is the solution obtained by rewriting the equation, using a substitution, and solving the resulting linear differential equation.
92 Medium
92 of 452
Application +4 | -1
The solution of the differential equation (x2−yx2)dydx+y2+xy2=0\left( {{x^2} - y{x^2}} \right)\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0 is
View answer and explanation
Correct option: A
log⁡(xy)=1x+1y+c\log \left( {\frac{x}{y}} \right) = \frac{1}{x} + \frac{1}{y} + c
Option A (log⁡(xy)=1x+1y+c\log \left( {\frac{x}{y}} \right) = \frac{1}{x} + \frac{1}{y} + c) is correct because it represents the solution obtained by separating variables and integrating the given differential equation.
93 Medium
93 of 452
Application +4 | -1
The solution of the differential equation (2y−1)dx−(2x+3)dy=0\left( {2y - 1} \right)dx - \left( {2x + 3} \right)dy = 0, is
View answer and explanation
Correct option: B
2x+32y−1=C\frac{{2x + 3}}{{2y - 1}} = C
Option B (2x+32y−1=C\frac{{2x + 3}}{{2y - 1}} = C) is correct because it is the general solution obtained by separating variables and integrating the given differential equation.
94 Medium
94 of 452
Application +4 | -1
The general solution of the differential equation (x−y2)dx+y(5x+y2)dy=0\left(x-y^2\right) d x+y\left(5 x+y^2\right) d y=0 is :
View answer and explanation
Correct option: A
(y2+x)4=C∣(y2+2x)3∣\left(y^2+x\right)^4=C\left|\left(y^2+2 x\right)^3\right|
Option B ((y2+2x)4=C∣(y2+x)3∣\left(y^2+2 x\right)^4=C\left|\left(y^2+x\right)^3\right|) is incorrect. The correct answer is A. This option has the terms switched in the final equation.
95 Medium
95 of 452
Application +4 | -1
If a curve passing through (1,1)(1,1) satisfies the differential equation x2dy=y2dx+xyd(xy)x^2 d y=y^2 d x+x y d(x y), then the equation of the curve is
View answer and explanation
Correct option: B
log⁡∣xy∣=1x−1y\log |x y|=\frac{1}{x}-\frac{1}{y}
Option B (log⁡∣xy∣=1x−1y\log |x y|=\frac{1}{x}-\frac{1}{y}) is correct because it represents the solution to the given differential equation after integration and applying the initial condition.
96 Medium
96 of 452
Application +4 | -1
General solution of the differential equation (xlog⁡x)dydx+y=2xlog⁡x(x \log x) \frac{d y}{d x}+y=\frac{2}{x} \log x is
View answer and explanation
Correct option: D
xylog⁡x=cx−2(1+log⁡∣x∣)x y \log x=c x-2(1+\log |x|)
Option D (xylog⁡x=cx−2(1+log⁡∣x∣)x y \log x=c x-2(1+\log |x|)) is correct because it is the solution obtained after solving the given linear differential equation.
97 Medium
97 of 452
Application +4 | -1
The solution of the differential equation ydx+xdyydx−xdy=x2exyy4\frac{y d x+x d y}{y d x-x d y}=\frac{x^2 e^{x y}}{y^4} satisfying y(0)=1y(0)=1, is
View answer and explanation
Correct option: B
x3=3y3(1−e−xy)x^3=3 y^3\left(1-e^{-x y}\right)
Option B (x3=3y3(1−e−xy)x^3=3 y^3(1-e^{-x y})) is correct because it is the solution to the differential equation that satisfies the initial condition y(0)=1.
98 Medium
98 of 452
Application +4 | -1
If y(t)y\left( t \right) is a solution of (1+t)dydt−ty=1 \left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1\, and y(0)=−1y\left( 0 \right) = - 1, then y(1)y\left( 1 \right) is equal to
View answer and explanation
Correct option: A
−12- \frac{1}{2}
Option A (−12\frac{{ - 1}}{2}) is correct because it is the value of y(1) obtained by solving the given differential equation and using the initial condition.
99 Medium
99 of 452
Application +4 | -1
The solution of the differential equation xdydx+y=xcos⁡x+sin⁡xx\frac{{dy}}{{dx}} + y = x\cos x + \sin x, given that y=1y = 1 when x=π2x = \frac{\pi }{2}, is
View answer and explanation
Correct option: C
y=sin⁡xy = \sin x
Option C (y=sin⁡xy = \sin x) is correct because it satisfies the given differential equation and the initial condition y=1y = 1 when x=π2x = \frac{\pi }{2}.
100 Medium
100 of 452
Application +4 | -1
The differential equation representing the family of ellipses having foci either on the xx-axis or on the yy-axis, centre at the origin and passing through the point (0,3)(0,3) is :
View answer and explanation
Correct option: A
xyy′−y2+9=0x y y^{\prime}-y^2+9=0
Option A (xyy′−y2+9=0x y y^{\prime}-y^2+9=0) is correct because it represents the differential equation of the family of ellipses passing through (0,3) with foci on either axis and center at the origin.
101 Medium
101 of 452
Application +4 | -1
The solution of the differential equation (x+y−1)dx−(2x+2y−3)dy=0(x+y-1) d x-(2 x+2 y-3) d y=0 is ……\ldots \ldots..
View answer and explanation
Correct option: C
2y+x+log⁡(x−y−2)−c2 y+x+\log (x-y-2)-c
Option C (2y+x+log⁡(x−y−2)−c2 y+x+\log (x-y-2)-c) is the correct solution to the given differential equation after applying appropriate substitutions and integration.
102 Medium
102 of 452
Analyzing +4 | -1
The order of the differential equation associated with the primitive y=c1+c2ex+c3e−2x+c4y = {c_1} + {c_2}{e^x} + {c_3}{e^{ - 2x + {c_4}}}, where c1,c2,c3,c4{c_1},{c_2},{c_3},{c_4} are arbitrary constants, is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the primitive can be simplified to have three arbitrary constants, thus the order of the differential equation is 3.
103 Medium
103 of 452
Analyzing +4 | -1
Let αx=exp⁡(xβyγ)\alpha x=\exp \left(x^{\beta} y^{\gamma}\right) be the solution of differential equation 2x2ydy−(1−xy2)dx=0,x>02 x^{2} y d y-\left(1-x y^{2}\right) d x=0, x>0, y(2)=log⁡e2y(2)=\sqrt{\log _{e} 2}. Then α+β−γ\alpha+\beta-\gamma equals
View answer and explanation
Correct option: B
1
Option B is correct. Solving the differential equation and using the given condition, we find α=2,β=1,γ=2\alpha=2, \beta=1, \gamma=2, thus α+β−γ=1\alpha+\beta-\gamma = 1.
104 Medium
104 of 452
Application +4 | -1
Solution of the differential equation dydx+ysec⁡2x=tan⁡xsec⁡2x\frac{d y}{d x}+y \sec ^2 x=\tan x \sec ^2 x is
View answer and explanation
Correct option: A
y=tan⁡x−1+ce−tan⁡xy=\tan x-1+c e^{-\tan x}
Option A (y=tan⁡x−1+ce−tan⁡xy=\tan x-1+c e^{-\tan x}) is correct because it is the solution obtained after finding the integrating factor and solving the linear differential equation.
105 Medium
105 of 452
Application +4 | -1
Solution of the differential equation dydx+yx=sin⁡x\frac{{dy}}{{dx}} + \frac{y}{x} = \sin x is
View answer and explanation
Correct option: A
x(y+cos⁡x)=sin⁡x+cx\left( {y + \cos x} \right) = \sin x + c
Option A (x(y+cos⁡x)=sin⁡x+cx\left( {y + \cos x} \right) = \sin x + c) is correct because it is the solution of the given linear differential equation after finding the integrating factor and integrating.
106 Medium
106 of 452
Application +4 | -1
The curve such that the initial ordinate of any tangent is less than the abscissa of the point of tangency by two unit is
View answer and explanation
Correct option: D
y=Cx−xlog⁡∣x∣−2y=C x-x \log |x|-2
The correct answer is D because the problem describes a first-order linear differential equation, which, when solved with the given condition, yields y=Cx-xlog|x|-2.
107 Medium
107 of 452
Application +4 | -1
If y=y(x)y=y(x) is the solution curve of the differential equation dydx+ytan⁡x=xsec⁡x,0≤x≤π3,y(0)=1\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1, then y(π6)y\left(\frac{\pi}{6}\right) is equal to
View answer and explanation
Correct option: D
π12−32log⁡e(2e3)\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)
The correct answer is D (π12−32log⁡e(2e3)\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)) because it is the correct solution to the given differential equation with the specified initial condition.
108 Medium
108 of 452
Application +4 | -1
Form of the differential equation of all family of lines y=mx+4my = mx + \frac{4}{m} by eliminating the arbitrary constant mm is
View answer and explanation
Correct option: B
x(dydx)2−ydydx+4=0x{\left( {\frac{{dy}}{{dx}}} \right)^2} - y\frac{{dy}}{{dx}} + 4 = 0
Option B (x(dydx)2−ydydx+4=0x{\left( {\frac{{dy}}{{dx}}} \right)^2} - y\frac{{dy}}{{dx}} + 4 = 0) is correct because it represents the differential equation of the given family of lines after eliminating the arbitrary constant m.
109 Medium
109 of 452
Application +4 | -1
If dydt+α⋅y=γ⋅e−βt\frac{d y}{d t}+\alpha \cdot y=\gamma \cdot e^{-\beta t}, then lim⁡t→∞y(t)\lim\limits_{t \to \infty } y\left( t \right) where α>0,β>0,γ>0,α≠β\alpha>0, \beta>0, \gamma>0, \alpha \neq \beta, is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because as t approaches infinity, the terms with negative exponents approach zero, leading to a limit of 0.
110 Medium
110 of 452
Application +4 | -1
A curve y=f(x)y=f(x) is passing through (0,0)(0,0). If the slope of the curve at any point (x,y)(x, y) is equal to (x+xy)(x+x y), then the number of solution of the equation f(x)=1f(x)=1, is:
View answer and explanation
Correct option: C
2
Option C (2) is correct because solving the differential equation and setting f(x)=1 leads to a quadratic equation with two real solutions.
111 Medium
111 of 452
Application +4 | -1
The solution of the differential equation dydx=e3x−2y+x2e−2y\frac{{dy}}{{dx}} = {e^{3x - 2y}} + {x^2}{e^{ - 2y}}, is
View answer and explanation
Correct option: B
12e2y=13(e3x+x3)+c\frac{1}{2}{e^{2y}} = \frac{1}{3}\left( {{e^{3x}} + {x^3}} \right) + c
Option B (12e2y=13(e3x+x3)+c\frac{1}{2}{e^{2y}} = \frac{1}{3}\left( {{e^{3x}} + {x^3}} \right) + c) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
112 Medium
112 of 452
Application +4 | -1
If dydx=x+y−2x−y\frac{d y}{d x}=\frac{x+y-2}{x-y} and y(0)=2y(0)=2, find y(2)y(2)
View answer and explanation
Correct option: A
0
Option A (0) is correct because solving the differential equation with the given initial condition leads to the solution where y(2) = 0.
113 Medium
113 of 452
Application +4 | -1
Solution of the differential equation y2dx+(x2−xy+y2)dy=0y^2 d x+\left(x^2-x y+y^2\right) d y=0 is
View answer and explanation
Correct option: A
y=keTan⁡−1(y/x)y=k e^{\operatorname{Tan}^{-1}(y / x)}
Option B (y=kxeTan⁡−1(y/x)y=k x e^{\operatorname{Tan}^{-1}(y / x)}) is incorrect. Option A (y=keTan⁡−1(y/x)y=k e^{\operatorname{Tan}^{-1}(y / x)}) is correct because it is the solution obtained after solving the homogeneous differential equation using the substitution y=vxy=vx.
114 Medium
114 of 452
Application +4 | -1
The solution of the differential equation x2dydx=x2+xy+y2x^2 \frac{d y}{d x}=x^2+x y+y^2 is
View answer and explanation
Correct option: A
tan⁡−1(yx)=log⁡x+c\tan ^{-1}\left(\frac{y}{x}\right)=\log x+c
Option A (tan⁡−1(yx)=log⁡x+c\tan ^{-1}\left(\frac{y}{x}\right)=\log x+c) is correct because it is the solution obtained after solving the homogeneous differential equation using the substitution y=vxy=vx.
115 Medium
115 of 452
Application +4 | -1
The general solution of the differential equation dydx=(1+y2)xy(1+x2)\frac{{dy}}{{dx}} = \frac{{\left( {1 + {y^2}} \right)}}{{xy\left( {1 + {x^2}} \right)}} is
View answer and explanation
Correct option: B
(1+x2)(1+y2)=cx2\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right) = c{x^2}
The correct answer is (1+x2)(1+y2)=cx2\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right) = c{x^2} because it is the result of separating variables and integrating the given differential equation.
116 Medium
116 of 452
Application +4 | -1
If dydx=xyx2+y2,y(1)=1\frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}},y\left( 1 \right) = 1, then one of the values of x0{x_0} satisfying y(x0)=ey\left( {{x_0}} \right) = e is given by
View answer and explanation
Correct option: B
e3e\sqrt 3
Option B (e3e\sqrt 3) is the correct answer because it satisfies the given differential equation and initial condition when y(x0)=ey(x_0) = e.
117 Medium
117 of 452
Application +4 | -1
The solution of the differential equation xlog⁡xdydx+y=2log⁡xx \log x \frac{d y}{d x}+y=2 \log x is
View answer and explanation
Correct option: C
ylog⁡x=(log⁡x)2+cy \log x=(\log x)^2+c
Option C (ylog⁡x=(log⁡x)2+cy \log x=(\log x)^2+c) is correct because it represents the solution to the given differential equation after finding the integrating factor and integrating.
118 Medium
118 of 452
Application +4 | -1
The solution of the differential equation (x+y)(dx−dy)=dx+dy(x+y)(d x-d y)=d x+d y, is
View answer and explanation
Correct option: D
x+y=kex−yx+y=k e^{x-y}
Option D (x+y=kex−yx+y=k e^{x-y}) is the correct solution obtained by separating variables and integrating the given differential equation.
119 Medium
119 of 452
Application +4 | -1
The differential equation dydx=1−y2y\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} determine of circles with
View answer and explanation
Correct option: C
fixed radius 1 and variable centres along the xx-axis
Option C (fixed radius 1 and variable centres along the xx-axis) is correct because the solution to the differential equation represents circles with a radius of 1 and centers on the x-axis.
120 Medium
120 of 452
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (x−x3)dy=(y+yx2−3x4)dx,x>2\left(x-x^3\right) d y=\left(y+y x^2-3 x^4\right) d x, x>2. If y(3)=3y(3)=3, then y(4)y(4) is equal to
View answer and explanation
Correct option: C
12
Option C (12) is correct because by solving the given differential equation and using the initial condition y(3)=3, we find y(4) = 12.
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