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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Medium

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Showing 1-30 of 452 questions
1 Medium
1 of 452
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (3y2−5x2)ydx+2x(x2−y2)dy=0\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0 such that y(1)=1y(1)=1. Then ∣(y(2))3−12y(2)∣\left|(y(2))^{3}-12 y(2)\right| is equal to :
View answer and explanation
Correct option: C
32232 \sqrt{2}
Option C (32232 \sqrt{2}) is correct because it involves solving a homogeneous differential equation with the given initial condition and then evaluating the expression at the solution.
2 Medium
2 of 452
Analyzing +4 | -1
The degree and order of the differential equation whose solution is a parabola whose axis is xx-axis, are
View answer and explanation
Correct option: B
1,21,2
Option B (1,2) is correct because the order of the differential equation is 2 (second derivative) and the degree is 1 (highest power of the highest order derivative).
3 Medium
3 of 452
Application +4 | -1
If x2+y2=1{x^2} + {y^2} = 1, then(y′=dydx,y′′=d2ydx2)\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)
View answer and explanation
Correct option: B
yy′′+(y′)2+1=0yy'' + {\left( {y'} \right)^2} + 1 = 0
Option B (yy′′+(y′)2+1=0yy'' + {\left( {y'} \right)^2} + 1 = 0) is correct because it is the result of differentiating the given equation twice and eliminating x.
4 Medium
4 of 452
Application +4 | -1
If x=Acos⁡4t+Bsin⁡4t,x = A\cos 4t + B\sin 4t, then d2xdt2\frac{{{d^2}x}}{{d{t^2}}} is equal to
View answer and explanation
Correct option: A
−16x- 16x
Option A (−16x- 16x) is correct because it is the second derivative of the given function, obtained by differentiating twice with respect to t.
5 Medium
5 of 452
Application +4 | -1
The integrating factor of the differential equation dydx=ytan⁡x−y2sec⁡x\frac{d y}{d x}=y \tan x-y^2 \sec x, is
View answer and explanation
Correct option: B
sec⁡x\sec x
Option B (sec⁡x\sec x) is correct because it is the integrating factor of the transformed Bernoulli's differential equation.
6 Medium
6 of 452
Application +4 | -1
Ify1(x){y_1}\left( x \right) is a solution of the differential equation dydx+f(x)y=0,\frac{{dy}}{{dx}} + f(x)y = 0, then a solution of
differential equation dydx+f(x)y=r(x)\frac{{dy}}{{dx}} + f(x)y = r(x) is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the provided options do not match the general solution obtained by solving the given differential equation using an integrating factor.
7 Medium
7 of 452
Application +4 | -1
An integrating factor for the differential equation (1+y2)dx−(tan⁡−1y−x)dy=0\left(1+y^2\right) d x-\left(\tan ^{-1} y-x\right) d y=0
View answer and explanation
Correct option: B
etan⁡−1ye^{\tan ^{-1} y}
Option B (etan⁡−1ye^{\tan ^{-1} y}) is the correct integrating factor for the given differential equation, found by rearranging the equation and identifying P(y).
8 Medium
8 of 452
Application +4 | -1
If for the differential equation y′=y^{\prime}= yx+ϕ(xy)\frac{y}{x}+\phi\left(\frac{x}{y}\right) the general solution is y=xlog⁡∣Cx∣y=\frac{x}{\log |C x|} then ϕ(x/y)\phi(x / y) is given by
View answer and explanation
Correct option: D
−y2/x2-y^2 / x^2
Option D (−y2/x2-y^2 / x^2) is correct because it satisfies the given differential equation and general solution.
9 Medium
9 of 452
Application +4 | -1
The solution of the differential equation ydx+(x+x2y)dy=0y\mathrm{ }dx + \left( {x + {x^2}y} \right)dy = 0 is
View answer and explanation
Correct option: B
−1xy+log⁡y=c- \frac{1}{{xy}} + \log y = c
Option B (−1xy+log⁡y=c- \frac{1}{{xy}} + \log y = c) is correct because it is the solution obtained by rearranging the equation and integrating.
10 Medium
10 of 452
Analyzing +4 | -1
The differential equation of all ellipses centred at the origin is
View answer and explanation
Correct option: B
xyy2+xy12−yy1=0xy{y_2} + xy_1^2 - y{y_1} = 0
Option B (xyy2+xy12−yy1=0xy{y_2} + xy_1^2 - y{y_1} = 0) is correct because it represents the differential equation obtained after eliminating the arbitrary constants from the equation of an ellipse centered at the origin.
11 Medium
11 of 452
Remembering +4 | -1
The degree of the differential equation 2(d2ydx2)+3(dydx)2+4y3.=x2\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) + 3{\left( {\frac{{dy}}{{dx}}} \right)^2} + 4{y^{3.}} = x, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the degree of the highest order derivative in the differential equation is 1.
12 Medium
12 of 452
Application +4 | -1
dydx=yx+2d(yx)ϕ(yx)\frac{d y}{d x}=\frac{y}{x}+2 \frac{d\left(\frac{y}{x}\right)}{\phi\left(\frac{y}{x}\right)} has the solution .......
View answer and explanation
Correct option: C
ϕ(yx)=kx2\phi\left(\frac{y}{x}\right)=k x^2
Option C (ϕ(yx)=kx2\phi(\frac{{y}}{{x}}) = kx^2) is correct because it is the solution obtained by substituting u=yxu = \frac{{y}}{{x}} and integrating the resulting separable equation.
13 Medium
13 of 452
Analyzing +4 | -1
The differential equation of the family of circles passing through the fixed points (a,0)\left( {a,0} \right) and (−a,0)\left( { - a,0} \right) is
View answer and explanation
Correct option: C
y1(y2−x2+a2)+2xy=0{y_1}\left( {{y^2} - {x^2} + {a^2}} \right) + 2xy = 0
Option C (y1(y2−x2+a2)+2xy=0{y_1}\left( {{y^2} - {x^2} + {a^2}} \right) + 2xy = 0) is correct because it represents the differential equation of the family of circles passing through the points (a, 0) and (-a, 0) after eliminating the arbitrary constant.
14 Medium
14 of 452
Analyzing +4 | -1
The differential equation of all parabolas having their axis of symmetry coinciding with the axis of XX, is
View answer and explanation
Correct option: A
yd2ydx2+(dydx)2=0y\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0
Option A (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 of all parabolas with their axis of symmetry coinciding with the x-axis.
15 Medium
15 of 452
Application +4 | -1
The solution y′=1−2y−4x1+y+2xy^{\prime}=\frac{1-2 y-4 x}{1+y+2 x} is given by
View answer and explanation
Correct option: B
(y+12(y+2x)2)=x+C\left(y+\frac{1}{2}(y+2 x)^2\right)=x+C
The correct answer is B (y + 1/2(y+2x)2) = x+C because it is the solution obtained by substituting u = y + 2x and integrating the resulting separable equation.

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16 Medium
16 of 452
Application +4 | -1
The solution of the differential equation 2xdydx−y=0;y(1)=22 x \frac{d y}{d x} - y=0 ; y(1)=2 represents ...... .
View answer and explanation
Correct option: B
A parabola
Option B (A parabola) is correct because the solution to the differential equation is of the form y2=kxy^2 = kx, which represents a parabola.
17 Medium
17 of 452
Application +4 | -1
The solution of the differential equation xx2+y2dy=(yx2+y2−1)dx\frac{x}{{{x^2} + {y^2}}}dy = \left( {\frac{y}{{{x^2} + {y^2}}} - 1} \right)dx, is
View answer and explanation
Correct option: C
y=xtan⁡(C−x)y = x\tan \left( {C - x} \right)
Option C (y=xtan⁡(C−x)y = x\tan (C - x)) is correct because it is the solution obtained after integrating the given differential equation using trigonometric substitution.
18 Medium
18 of 452
Application +4 | -1
If ey+xy=e{e^y} + xy = e, the ordered pair (dydx,d2ydx2)\left( {\frac{{dy}}{{dx}},\frac{{{d^2}y}}{{d{x^2}}}} \right) at x=0 is equal to
View answer and explanation
Correct option: B
(−1e,1e2)\left( { - \frac{1}{e},\frac{1}{{{e^2}}}} \right)
Option B ((−1e,1e2)\left( { - \frac{{1}}{e},\frac{{1}}{{{{e}^2}}}} \right)) is correct because it is the ordered pair of the first and second derivatives of the given equation evaluated at x=0.
19 Medium
19 of 452
Analyzing +4 | -1
Number of straight lines which satisfy the differential equation dydx+x(dydx)2−y=0\frac{d y}{d x}+x\left(\frac{d y}{d x}\right)^2-y=0 is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the differential equation represents a family of straight lines, and solving for the constants yields two possible straight lines.
20 Medium
20 of 452
Application +4 | -1
The differential equation of all circles passing through the origin and having their centres on the xx-axis is
View answer and explanation
Correct option: C
y2=x2+2xydydx{y^2} = {x^2} + 2xy\frac{{dy}}{{dx}}
Option C (y2=x2+2xydydx{y^2} = {x^2} + 2xy\frac{{dy}}{{dx}}) is correct because it represents the differential equation derived from the equation of circles passing through the origin with centers on the x-axis.
21 Medium
21 of 452
Application +4 | -1
The differential equation y5x+y−xdydx=0y^5 x+y-x \frac{d y}{d x}=0 has the general solution ......
View answer and explanation
Correct option: B
x55+14(xy)4=c\frac{x^5}{5}+\frac{1}{4}\left(\frac{x}{y}\right)^4=c
Option B (x55+14(xy)4=c\frac{{x^5}}{{5}} + \frac{{1}}{{4}}(\frac{{x}}{{y}})^4 = c) is correct because it is the result of solving the given differential equation by using a suitable substitution and integration.
22 Medium
22 of 452
Application +4 | -1
The solution of the differential equation (e−2xx−yx)dxdy=1,x≠0\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}=1, x \neq 0 is ...... -
View answer and explanation
Correct option: A
ye2x=2x+cy e^{2 \sqrt{x}}=2 \sqrt{x}+c
Option A (ye2x=2x+cy e^{2 \sqrt{x}}=2 \sqrt{x}+c) is the correct solution, derived by using an integrating factor after rearranging the equation.
23 Medium
23 of 452
Application +4 | -1
The solution to the differential equation dydx=e3x−2y+x2e−2y\frac{d y}{d x}=e^{3 x-2 y}+x^2 e^{-2 y} is
View answer and explanation
Correct option: B
3e2y=2(e3x+x3)+c3 {e}^{2 {y}}=2\left({e}^{3 {x}}+{x}^3\right)+{c}
Option B (3e2y=2(e3x+x3)+c3 {e}^{2 {y}}=2\left({e}^{3 {x}}+{x}^3\right)+{c}) is correct because separating variables and integrating leads to the given solution.
24 Medium
24 of 452
Application +4 | -1
The solution of the differential equation dydx+2ycot⁡x=3x2cosec⁡2x\frac{d y}{d x}+2 y \cot x=3 x^2 \operatorname{cosec}^2 x is
View answer and explanation
Correct option: A
ysin⁡2x=x3+cy \sin ^2 x=x^3+c
Option A (ysin⁡2x=x3+cy \sin ^2 x=x^3+c) is correct because it is the solution to the given differential equation after finding the integrating factor and integrating.
25 Medium
25 of 452
Application +4 | -1
The solution of the differential equation x(x−y)dydx=y(x+y)x\left( {x - y} \right)\frac{{dy}}{{dx}} = y\left( {x + y} \right), is
View answer and explanation
Correct option: A
xy+log⁡(xy)=c\frac{x}{y} + \log \left( {xy} \right) = c
Option A (xy+log⁡(xy)=c\frac{x}{y} + \log \left( {xy} \right) = c) is correct because it is the solution obtained by substituting y=vxy = vx and integrating the resulting equation.
26 Medium
26 of 452
Application +4 | -1
The solution of the differential equation dydx=(4x+y+1)2\frac{{dy}}{{dx}} = {\left( {4x + y + 1} \right)^2}, is
View answer and explanation
Correct option: D
(4x+y+1)=2tan⁡(2x+c)\left( {4x + y + 1} \right) = 2\tan \left( {2x + c} \right)
Option D ((4x+y+1)=2tan⁡(2x+c)(4x + y + 1) = 2\tan (2x + c)) is correct because it is the solution of the given differential equation after substitution and integration.
27 Medium
27 of 452
Application +4 | -1
If y(0)=1y(0)=1, then the solution of the differential equation x3dydx−x2y=−y4cos⁡xx^3 \frac{d y}{d x}-x^2 y=-y^4 \cos x is given by
View answer and explanation
Correct option: B
x3=3y3sin⁡xx^3=3 y^3 \sin x
Option B (x3=3y3sin⁡xx^3=3 y^3 \sin x) is the correct solution to the given differential equation, obtained by transforming it into a linear form and applying the initial condition.
28 Medium
28 of 452
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.
29 Medium
29 of 452
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.
30 Medium
30 of 452
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.
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