Skip to main content
Back to DIFFERENTIAL EQUATIONS

Your Progress

0 / 975 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 661-690 of 975 questions

Preparing practice and saved progress…

661 Medium
661 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (2xlog⁡ex)dydx+2y=3xlog⁡ex,x>0\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x > 0 and y(e−1)=0y\left(e^{-1}\right)=0. Then, y(e)y(e) is equal to
View answer and explanation
Correct option: C
−3e-\frac{3}{\mathrm{e}}
Option C (-3/e) is correct because it is the value of y(e) obtained by solving the given differential equation with the given initial condition.
662 Medium
662 of 975
Analyzing +4 | -1
The order of the differential equation whose general solution gives by y=(c1+c2)cos⁡(x+c3)−c4ex+c5y=\left(c_1+c_2\right) \cos \left(x+c_3\right)-c_4 e^{x+c_5} where c1,c2,c3,c4,c5\mathrm{c}_1, \mathrm{c}_2, \mathrm{c}_3, \mathrm{c}_4, \mathrm{c}_5 are arbitrary constants is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the general solution can be simplified to have only three independent arbitrary constants.
663 Easy
663 of 975
Application +4 | -1
dydx+ytan⁡x=sec⁡x\frac{d y}{d x}+y \tan x=\sec x has the solution ........
View answer and explanation
Correct option: A
y=sin⁡x+ccos⁡xy=\sin x+c \cos x
Option A (y=sin⁡x+ccos⁡xy = \sin x + c \cos x) is correct because it is the solution obtained by solving the linear differential equation using an integrating factor.
664 Medium
664 of 975
Application +4 | -1
The solution of x3dydx+4x2tan⁡y=exx^3 \frac{d y}{d x}+4 x^2 \tan y=e^x sec yy satisfying y(1)=0y(1)=0 is
View answer and explanation
Correct option: C
tan⁡y=(x−1)exx−3\tan y=(x-1) e^x x^{-3}
Option D (sin⁡y=ex(x−1)x−3\sin y=e^x(x-1) x^{-3}) is correct because it satisfies the given differential equation and the initial condition y(1)=0y(1)=0.
665 Easy
665 of 975
Application +4 | -1
The general solution of a differential equation dydx=ex+y\frac{d y}{d x}=e^{x+y} is :
View answer and explanation
Correct option: A
ex+e−y=Ce^x+e^{-y}=C
Option A (ex+e−y=Ce^x+e^{-y}=C) is correct because separating variables and integrating leads to this solution.
666 Hard
666 of 975
Application +4 | -1
Solution of the equation xdy=(y+xf(y/x)f′(y/x))dxx d y=\left(y+\frac{x f(y / x)}{f^{\prime}(y / x)}\right) d x is -
View answer and explanation
Correct option: C
∣f(y/x)∣=c∣x∣,c>0|f(y / x)|=c|x|, c>0
Option C (∣f(y/x)∣=c∣x∣,c>0|f(y / x)|=c|x|, c>0) is correct because it represents the general solution to the given homogeneous differential equation.
667 Medium
667 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (1+x2)dydx+y=etan⁡−1x,y(1)=0\left(1+x^2\right) \frac{d y}{d x}+y=\mathrm{e}^{\tan ^{-1} x}, y(1)=0. Then y(0)\mathrm{y}(0) is
View answer and explanation
Correct option: B
12(1−eπ/2)\frac{1}{2}\left(1-\mathrm{e}^{\pi / 2}\right)
Option A (12(1−eπ/2)\frac{{1}}{{2}}(1 - e^{\pi/2})) is correct because it is the value of y(0) obtained by solving the given differential equation with the given initial condition.
668 Easy
668 of 975
Application +4 | -1
The Solution of 1−x2sin⁡−1xdy−ydx=0\sqrt{1-x^2} \sin ^{-1} x d y-y d x=0
View answer and explanation
Correct option: A
y=csin⁡−1xy=c \sin ^{-1} x
Option A (y=csin⁡−1xy=c \sin ^{-1} x) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
669 Medium
669 of 975
Application +4 | -1
If the differential equation dydx=yx+ϕ(xy)\frac{d y}{d x}=\frac{y}{x}+\phi\left(\frac{x}{y}\right) has the general solution y=xlog⁡∣cxy=\frac{x}{\log \mid c x}, then ϕ(xy)=……\phi\left(\frac{x}{y}\right)=\ldots \ldots
View answer and explanation
Correct option: D
−y2x2\frac{-y^2}{x^2}
Option D (−y2x2\frac{{-y^2}}{{x^2}}) is correct because it satisfies the given differential equation and general solution after substitution and differentiation.
670 Easy
670 of 975
Analyzing +4 | -1
Let ff be a differentiable function such that x2f(x)−x=4∫0xtf(t)dt,f(1)=23x^2 f(x)-x=4 \int\limits_{0}^x t f(t) d t, f(1)=\frac{2}{3}. Then 18f(3)18 f(3) is equal to
View answer and explanation
Correct option: B
160
Option B (160) is correct because after solving the differential equation and applying the initial condition, we find f(3) and then calculate 18f(3).
671 Medium
671 of 975
Application +4 | -1
Solution of the differential equation dydxtan⁡y=sin⁡(x+y)+sin⁡(x−y)\frac{{dy}}{{dx}}\tan y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right) is
View answer and explanation
Correct option: A
sec⁡y+2cos⁡x=c\sec y + 2\cos x = c
Option A (sec⁡y+2cos⁡x=c\sec y + 2\cos x = c) is correct because it is the solution to the given differential equation.
672 Easy
672 of 975
Application +4 | -1
If aa is an arbitrary constant, then solution of the differential equation dydx+1−y21−x2=0\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0, is
View answer and explanation
Correct option: A
x1−y2+y1−x2=ax \sqrt{1-y^2}+y \sqrt{1-x^2}=a
Option A (x1−y2+y1−x2=ax \sqrt{1-y^2}+y \sqrt{1-x^2}=a) is the correct solution obtained by separating variables and using trigonometric identities after integration.
673 Medium
673 of 975
Application +4 | -1
The solution of dydx+1=ex+y\frac{{dy}}{{dx}} + 1 = {e^{x + y}} is
View answer and explanation
Correct option: A
e−(x+y)+x+c=0{e^{ - \left( {x + y} \right)}} + x + c = 0
Option A (e−(x+y)+x+c=0{e^{ - \left( {x + y} \right)}} + x + c = 0) is correct because it is the solution obtained by solving the given differential equation using a substitution.
674 Medium
674 of 975
Application +4 | -1
y=cx−c2y = cx - {c^2}, is the general solution of the differential equation
View answer and explanation
Correct option: A
(y′)2−xy′+y=0{\left( {y'} \right)^2} - xy' + y = 0
Option A ((y′)2−xy′+y=0{\left( {y'} \right)^2} - xy' + y = 0) is correct because it is obtained by eliminating the arbitrary constant 'c' from the given general solution.
675 Medium
675 of 975
Application +4 | -1
The solution of y′+y=−xyy^{\prime}+y=-\frac{x}{y} is given by
View answer and explanation
Correct option: C
y2=−x+12(1+e−2π)y^2=-x+\frac{1}{2}\left(1+e^{-2 \pi}\right)
The correct answer is C (y2=−x+12(1+e−2π)y^2=-x+\frac{{1}}{{2}}\left(1+e^{-2 \pi}\right)) because it is the solution of the given differential equation with the initial condition y(0)=1.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
676 Easy
676 of 975
Application +4 | -1
The integrating factor of the differential equation (x2−1)dydx+2xy=x\left(x^2-1\right) \frac{d y}{d x}+2 x y=x is
View answer and explanation
Correct option: B
x2−1x^2-1
Option B (x2-1) is correct because the integrating factor for the given differential equation is x2-1.
677 Medium
677 of 975
Application +4 | -1
Differential equation of y=sec⁡(tan⁡−1x)y = \sec ({\tan ^{ - 1}}x) is
View answer and explanation
Correct option: C
(1+x2)dydx=xy\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = xy
Option C ((1+x2)dydx=xy\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} = xy) is correct because it is the differential equation derived from the given equation.
678 Medium
678 of 975
Application +4 | -1
If dydx=cos⁡(x+y),y(π2)=0\frac{d y}{d x}=\cos (x+y), y\left(\frac{\pi}{2}\right)=0, then y(0)=y(0)=
View answer and explanation
Correct option: D
−2tan⁡−1(π2−1)-2 \tan ^{-1}\left(\frac{\pi}{2}-1\right)
Option D (−2tan⁡−1(π2−1)-2 \tan ^{-1}\left(\frac{\pi}{2}-1\right)) is correct because it is the solution obtained after solving the given differential equation with the initial condition.
679 Medium
679 of 975
Understanding +4 | -1
The differential equation of all straight lines passing through origin is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the differential equation represents the family of straight lines passing through the origin is y=xdydxy = x \frac{{dy}}{{dx}}.
680 Easy
680 of 975
Application +4 | -1
If x=sin⁡t,y=cos⁡ptx = \sin t,y = \cos pt, then
View answer and explanation
Correct option: D
(1−x2)y2−xy1+p2y=0\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0
Option D ((1−x2)y2−xy1+p2y=0\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0) is correct because it is the differential equation obtained by differentiating the given parametric equations.
681 Medium
681 of 975
Application +4 | -1
The differential equations of all circles passing through origin and having their centres on the x\mathrm{x}-axis is
View answer and explanation
Correct option: D
dydx=y2−x22xy\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y^2-x^2}{2 x y}
Option D (dydx=y2−x22xy\frac{{\mathrm{{dy}}}}{{\mathrm{{dx}}}} = \frac{{y^2-x^2}}{{2 x y}}) is correct because it represents the differential equation of all circles passing through the origin and having their centers on the x-axis.
682 Medium
682 of 975
Application +4 | -1
The solution of the differential equation sec⁡2xtan⁡ydx+sec⁡2ytan⁡xdy=0{\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0 is
View answer and explanation
Correct option: A
tan⁡ytan⁡x=c\tan y\tan x = c
Option A (tan⁡ytan⁡x=c\tan y\tan x = c) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
683 Easy
683 of 975
Understanding +4 | -1
d3ydx3+2[1+d2ydx2]=1\frac{d^3 y}{d x^3}+2\left[1+\frac{d^2 y}{d x^2}\right]=1 has degree and order is :
View answer and explanation
Correct option: A
1,3
Option A (1,3) is correct because the order is the highest derivative (3) and the degree is the power of the highest derivative (1).
684 Medium
684 of 975
Application +4 | -1
The solution of the differential equation dydxtan⁡y=sin⁡(x+y)+sin⁡(x−y)\frac{{dy}}{{dx}}\tan y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right) is
View answer and explanation
Correct option: A
sec⁡y+2cos⁡x=c\sec y + 2\cos x = c
Option A (sec⁡y+2cos⁡x=c\sec y + 2\cos x = c) is correct because it is the result of integrating the given differential equation after separation of variables and using trigonometric identities.
685 Medium
685 of 975
Application +4 | -1
The solution of dy=cos⁡x(2−ycosecx)dxdy = \cos x\left( {2 - y\mathrm{cosec}x} \right)dx, where y=2y = \sqrt 2, when x=π4x = \frac{\pi }{4} is
View answer and explanation
Correct option: A
y=sin⁡x+12cosecxy = \sin x + \frac{1}{2}\mathrm{cosec}x
Option A (y=sin⁡x+12cosecxy = \sin x + \frac{1}{2}\mathrm{cosec}x) is correct because it satisfies the given differential equation and the initial condition y=2y = \sqrt 2 when x=π4x = \frac{\pi }{4}.
686 Medium
686 of 975
Application +4 | -1
The solution of dydx=ax+gby+f\frac{{dy}}{{dx}} = \frac{{ax + g}}{{by + f}} represents a circle when
View answer and explanation
Correct option: B
a=−ba = - b
Option B (a=−ba = - b) is correct because the given differential equation represents a circle only when the coefficients of x2x^2 and y2y^2 are equal in magnitude and opposite in sign.
687 Medium
687 of 975
Application +4 | -1
The motion of a particle is given by dxdt=\frac{d x}{d t}= −3acos⁡2tsin⁡t,dydt=3asin⁡2tcos⁡t-3 a \cos ^2 t \sin t, \frac{d y}{d t}=3 a \sin ^2 t \cos t and passes through at tt =0=0, then the path is given by
View answer and explanation
Correct option: D
(x+a)2/3+y2/3=a2/3(x+a)^{2 / 3}+y^{2 / 3}=a^{2 / 3}
The correct answer is D ((x+a)2/3+y2/3=a2/3(x+a)^{2 / 3}+y^{2 / 3}=a^{2 / 3}) because it represents the path of the particle given the parametric equations and initial condition.
688 Easy
688 of 975
Understanding +4 | -1
A differential equation of the form dydx=F(x,y)\frac{d y}{d x}=F(x, y) is said to be homogeneous, if F(x,y){F}({x}, {y}) is a homogeneous function of degree,
View answer and explanation
Correct option: A
0
0 is correct because a homogeneous function F(x, y) of degree zero satisfies F(tx, ty) = F(x, y).
689 Easy
689 of 975
Application +4 | -1
If the length of the subnormal at any point of the curve is constant, then the eccentricity of this curve is
View answer and explanation
Correct option: D
e=1\mathrm{e}=1
Option D (e=1) is correct because the condition of constant subnormal implies the curve is a parabola, which has an eccentricity of 1.
690 Medium
690 of 975
Application +4 | -1
The solution of the equation dydx=x2y−x\frac{d y}{d x}=\frac{x}{2 y-x} is
View answer and explanation
Correct option: A
(x−y)(x+2y)2=c(x-y)(x+2 y)^2=c
Option A ((x−y)(x+2y)2=c(x-y)(x+2 y)^2=c) is the correct solution to the differential equation, which is solved by substituting y=vxy=vx.
Revision notes and key takeaways

Review DIFFERENTIAL EQUATIONS in your textbook or class notes, then use this set to check your understanding.

Showing 661-690 of 975 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - DIFFERENTIAL EQUATIONS MCQs (Easy) 494 MCQs → ENGINEERING MATHEMATICS - DIFFERENTIAL EQUATIONS MCQs (Hard) 29 MCQs → ENGINEERING MATHEMATICS - DIFFERENTIAL EQUATIONS MCQs (Medium) 452 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

Your practice

0 / 975 answered · 975 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →