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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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Showing 1-30 of 975 questions
1 Medium
1 of 975
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 975
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 975
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 975
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 Hard
5 of 975
Application +4 | -1
Let y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) be the solution curves of the differential equation dydx=y+7\frac{d y}{d x}=y+7 with initial conditions y1(0)=0y_{1}(0)=0 and y2(0)=1y_{2}(0)=1 respectively. Then the curves y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) intersect at
View answer and explanation
Correct option: A
No point
Option A (No point) is correct because solving the differential equation and applying the initial conditions leads to two lines that never intersect.
6 Easy
6 of 975
Application +4 | -1
Let y=y(t)y=y(t) be a solution of the differential equation dydt+αy=γe−βt\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}
where, α>0,β>0\alpha>0, \beta>0 and γ>0\gamma>0. Then lim⁡x→∞y(t)\lim \limits_{{x \rightarrow \infty}} y(t)
View answer and explanation
Correct option: B
Is 0
Option B (Is 0) is correct because as t approaches infinity, both terms in the general solution of the differential equation approach 0 due to the negative exponential terms.
7 Medium
7 of 975
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.
8 Easy
8 of 975
Understanding +4 | -1
The order and degree of the following differential equation [1+(dydx)2]52=d3ydx3{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{\frac{5}{2}}} = \frac{{{d^3}y}}{{d{x^3}}} are respectively
View answer and explanation
Correct option: A
3,23,2
Option A (3,2) is correct because the order is 3 (third derivative) and after squaring, the degree of the third derivative is 2.
9 Easy
9 of 975
Application +4 | -1
Solution of the differential equation yex/ydx=(xex/y+y2sin⁡y)dyy e^{x / y} d x=\left(x e^{x / y}+y^2 \sin y\right) d y, is
View answer and explanation
Correct option: D
ex/y=−cos⁡y+Ce^{x / y}=-\cos y+C
Option D (ex/y=−cos⁡y+Ce^{x / y}=-\cos y+C) is the correct solution obtained by recognizing the exact form and integrating the given differential equation.
10 Medium
10 of 975
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.
11 Easy
11 of 975
Analyzing +4 | -1
The order of the differential equation whose general solution is given by
y=(c1+c2)cos⁡(x+c3)−c4ex+c5y = \left( {{c_1} + {c_2}} \right)\cos (x + {c_3}) - {c_4}{e^{x + {c_5}}} where c1,c2,c3,c4,c5{c_1},{c_2},{c_3},{c_4},{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 contain only three independent arbitrary constants, thus the order of the differential equation is 3.
12 Medium
12 of 975
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).
13 Easy
13 of 975
Application +4 | -1
The curve y=f(x)(f(x)≥0,f(0)=0)y=f(x)(f(x) \geq 0, f(0)=0) bounding a curvilinear trapezoid with the base [0,x][0, x], whose area is proportional to the (n+1)(n+1) th power of f(x)f(1)=1f(x) f(1)=1 is given by
View answer and explanation
Correct option: A
x=ynx=y^n
Option A (x=ynx=y^n) is correct because it satisfies the given condition that the area under the curve is proportional to the (n+1)th power of f(x).
14 Easy
14 of 975
Application +4 | -1
The solution of differential equation tan⁡ysec⁡2xdx+tan⁡xsec⁡2ydy=0\tan y \sec ^2 x d x+\tan x \sec ^2 y d y=0 is
View answer and explanation
Correct option: D
tan⁡x⋅tan⁡y=k\tan x \cdot \tan y=k
Option D (tan⁡x⋅tan⁡y=k\tan x \cdot \tan y=k) is correct because separating variables and integrating leads to this solution.
15 Easy
15 of 975
Understanding +4 | -1
The degree of the differential equation (d3ydx3)2/3+4−3d2ydx2+5dydx=0\left(\frac{d^3 y}{d x^3}\right)^{2 / 3}+4-3 \frac{d^2 y}{d x^2}+5 \frac{d y}{d x}=0, is
View answer and explanation
Correct option: B
2
Option B is correct because after raising both sides to the power of 3 to eliminate the fractional exponent, the highest order derivative is d3y/dx3, and its power is 2, hence the degree is 2.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
16 Easy
16 of 975
Application +4 | -1
Solution of the differential equation xdydx=y+x2+y2x \frac{d y}{d x}=y+\sqrt{x^2+y^2}, is
View answer and explanation
Correct option: D
y+x2+y2=Cx2y+\sqrt{x^2+y^2}=C x^2
The correct answer is D (y+x2+y2=Cx2y+\sqrt{x^2+y^2}=C x^2) because it is the solution obtained by solving the given homogeneous differential equation using the substitution y=vxy=vx.
17 Easy
17 of 975
Application +4 | -1
Solution of differential equation dydx+yx=x2y6\frac{d y}{d x}+\frac{y}{x}=x^2 y^6 is of form 2x5y5=Ax2+C\frac{2}{x^5 y^5}=\frac{A}{x^2}+C. Then find value of AA
View answer and explanation
Correct option: A
5
Option A (5) is correct because by solving the differential equation and comparing with the given form, the value of A is found to be 5.
18 Easy
18 of 975
Application +4 | -1
A differential equation representing the family of parabolas with axis parallel to YY-axis and whose length of latus rectum is the distance of the point (2, -3) from the line is given by 3x+4y=53x + 4y = 5, is given by
View answer and explanation
Correct option: D
11d2ydx2=1011\frac{{{d^2}y}}{{d{x^2}}} = 10
Option D (11d2ydx2=1011\frac{{{d^2}y}}{{d{x^2}}} = 10) is correct because it represents the differential equation derived from the given family of parabolas after eliminating the arbitrary constants.
19 Medium
19 of 975
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.
20 Easy
20 of 975
Understanding +4 | -1
If pp and qq are order and degree of differential equation y2(d2ydx2)2+3x(dydx)1/3+x2y2=sin⁡xy^2\left(\frac{d^2 y}{d x^2}\right)^2+3 x\left(\frac{d y}{d x}\right)^{1 / 3}+x^2 y^2=\sin x, then:
View answer and explanation
Correct option: D
p+q=8p+q=8
Option A (a periodic function with period 4a4a) is correct because the function repeats its values after an interval of 4a4a, satisfying the definition of a periodic function.
21 Easy
21 of 975
Application +4 | -1
The curve that passes through the point (2,3)(2,3) and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by:
View answer and explanation
Correct option: A
y=6xy=\frac{6}{x}
Option A (y=6xy=\frac{{6}}{{x}}) is correct because it satisfies the given geometric property and passes through the point (2,3).
22 Medium
22 of 975
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.
23 Medium
23 of 975
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.
24 Medium
24 of 975
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.
25 Medium
25 of 975
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.
26 Easy
26 of 975
Application +4 | -1
If x(t)x(t) is a solution of (t+1)dx=(2x+(t+1)4)(t+1) d x=\left(2 x+(t+1)^4\right) dt,x(0)=2d t, x(0)=2 then lim⁡t→1x(t)\lim \limits_{{t \rightarrow 1}} x(t) is
View answer and explanation
Correct option: C
14
Option C (14) is correct because it is the limit of the solution to the differential equation as t approaches 1, given the initial condition.
27 Easy
27 of 975
Application +4 | -1
If y(t)y(t) is a solution of (1+t)dydt−ty=1(1+t) \frac{d y}{d t}-t y=1 and y(0)=1y(0)=1, then y(1)y(1) is equal to
View answer and explanation
Correct option: A
−12-\frac{1}{2}
Option A (y=xsec⁡xy=x \sec x) is correct because it is the solution of the given differential equation with the initial condition y(0) = 0.
28 Easy
28 of 975
Application +4 | -1
Which of the following is a homogeneous differential equations?
View answer and explanation
Correct option: D
xsin⁡(yx)dy={ysin⁡(yx)−x}dxx \sin \left(\frac{y}{x}\right) d y=\left\{y \sin \left(\frac{y}{x}\right)-x\right\} d x
Option D (xsin⁡(yx)dy={ysin⁡(yx)−x}dxx \sin \left(\frac{{y}}{{x}}\right) d y=\left\{y \sin \left(\frac{{y}}{{x}}\right)-x\right\} d x) is correct because it can be written in the form dy/dx = f(y/x), which is the definition of a homogeneous differential equation.
29 Hard
29 of 975
Application +4 | -1
If y′′−3y′+2y=0y'' - 3y' + 2y = 0 where y(0)=1,y′(0)=0y\left( 0 \right) = 1,y'\left( 0 \right) = 0, then the value of yy at x=log⁡2x = \log 2 is
View answer and explanation
Correct option: D
0
Option D (0) is correct because substituting x=log⁡2x = \log 2 into the solution of the differential equation, obtained using the given initial conditions, yields y=0y = 0.
30 Medium
30 of 975
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.
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