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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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811 Medium
811 of 975
Application +4 | -1
The differential equation of all circles which passes through the origin and whose centre lies on yy-axis, is
View answer and explanation
Correct option: A
(x2−y2)dydx−2xy=0\left( {{x^2} - {y^2}} \right)\frac{{dy}}{{dx}} - 2xy = 0
Option A ((x2−y2)dydx−2xy=0\left( {{x^2} - {y^2}} \right)\frac{{dy}}{{dx}} - 2xy = 0) is correct because it represents the differential equation of all circles passing through the origin with centers on the y-axis.
812 Medium
812 of 975
Application +4 | -1
The curve passing through the point (0,π/4)(0, \pi / 4) satisfying the differential equation dydx+xsin⁡2y=x3cos⁡2y\frac{d y}{d x}+x \sin 2 y=x^3 \cos ^2 y is
View answer and explanation
Correct option: C
tan⁡y=12(x2−1)+32e−x2\tan y=\frac{1}{2}\left(x^2-1\right)+\frac{3}{2} e^{-x^2}
Option D (tan⁡y=12(x2−1)+32e−x2\tan y=\frac{1}{2}(x^2-1)+\frac{3}{2} e^{-x^2}) is the correct solution to the differential equation, obtained by transforming it into a linear form and applying the initial condition.
813 Easy
813 of 975
Analyzing +4 | -1
The solution of the differential equation y3x2cos⁡(x3)dx+sin⁡(x3)y2dy=x3dxy^3 x^2 \cos \left(x^3\right) d x+\sin \left(x^3\right) y^2 d y=\frac{x}{3} d x is 2sin⁡(x3)yk=x2+C2 \sin \left(x^3\right) y^k=x^2+C (where C{C} is an arbitrary constant) then the value of k{k} is equal to
View answer and explanation
Correct option: A
3
Option A (3) is correct because by recognizing the exact differential and integrating, the value of k in the solution is found to be 3.
814 Medium
814 of 975
Analyzing +4 | -1
A continuously differential function ϕ(x)\phi \left( x \right) in (0,π)\left( {0,\pi } \right) satisfying y′=1+y2,y(0)=0=y(π)y' = 1 + {y^2},y\left( 0 \right) = 0 = y\left( \pi \right), is
View answer and explanation
Correct option: D
Not possible
Option D (Not possible) is correct because the solution to the differential equation is not continuous in the given interval.
815 Easy
815 of 975
Understanding +4 | -1
The order and degree of the differential equation sin⁡x(dx+dy)=cos⁡x(dx−dy)\sqrt {\sin x} \left( {dx + dy} \right) = \sqrt {\cos x} \left( {dx - dy} \right) are
View answer and explanation
Correct option: C
(1,1)\left( {1,1} \right)
Option C ((1,1)\left( {1,1} \right)) is correct because the highest order derivative is dydx\frac{{dy}}{{dx}} (order 1) and its power is 1 (degree 1).
816 Easy
816 of 975
Application +4 | -1
The solution of dydx=y2xy−x2\frac{{dy}}{{dx}} = \frac{{{y^2}}}{{xy - {x^2}}} is
View answer and explanation
Correct option: B
eyx=ky{e^{\frac{y}{x}}} = ky
Option B (eyx=ky{e^{\frac{y}{x}}} = ky) is correct because it is the solution of the given homogeneous differential equation after applying the appropriate substitution and integration.
817 Easy
817 of 975
Application +4 | -1
The solution of the differential equation dydx+yx=x2\frac{d y}{d x}+\frac{y}{x}=x^2 is
View answer and explanation
Correct option: A
4xy=x4+c4 x y=x^4+c
Option A (4xy=x4+c4 x y=x^4+c) is the correct solution, obtained by solving the linear differential equation using an integrating factor.
818 Medium
818 of 975
Application +4 | -1
Solution of the differential equation cos⁡xdy=y(sin⁡x−y)dx,0<x<π2\cos xdy = y\left( {\sin x - y} \right)dx,0 < x < \frac{\pi }{2}, is
View answer and explanation
Correct option: A
sec⁡x=(tan⁡x+c)y\sec x = \left( {\tan x + c} \right)y
Option A (sec⁡x=(tan⁡x+c)y\sec x = (\tan x + c)y) is correct because it represents the solution obtained after solving the given differential equation using an integrating factor.
819 Easy
819 of 975
Application +4 | -1
The general solution of the differential equation 100d2ydx2−20dydx+y=0100\frac{{{d^2}y}}{{d{x^2}}} - 20\frac{{dy}}{{dx}} + y = 0 is
View answer and explanation
Correct option: C
y=(c1+c2x)e x10y = \left( {{c_1} + {c_2}x} \right)e{\,^{\frac{x}{{10}}}}
Option C (y=(c1+c2x)e x10y = \left( {{{c_1} + {c_2}x}} \right)e{\,^{\frac{{x}}{{10}}}}) is correct because the auxiliary equation has repeated roots, leading to this form of the general solution for a second-order homogeneous differential equation.
820 Medium
820 of 975
Application +4 | -1
A particular solution of log⁡(dydx)=3x+4y,y(0)=0\log \left( {\frac{{dy}}{{dx}}} \right) = 3x + 4y,y\left( 0 \right) = 0 is
View answer and explanation
Correct option: C
3e3x+4e−4y=73{e^{3x}} + 4{e^{ - 4y}} = 7
Option D (4e3x+3e−4y=74{e^{3x}} + 3{e^{ - 4y}} = 7) is correct because it is the particular solution of the given differential equation that satisfies the initial condition y(0) = 0.
821 Medium
821 of 975
Application +4 | -1
The solution of dy dx=12x−y2\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2 x-y^2} is given by
View answer and explanation
Correct option: D
x=Ce2y+12y2+12y+14x=C e^{2 y}+\frac{1}{2} y^2+\frac{1}{2} y+\frac{1}{4}
Option D (x=Ce2y+12y2+12y+14x=C e^{2 y}+\frac{1}{2} y^2+\frac{1}{2} y+\frac{1}{4}) is correct because it is the solution to the differential equation dydx=12x−y2\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2 x-y^2}.
822 Medium
822 of 975
Application +4 | -1
Solution of the equation (x+log⁡y)dy+ydx=0(x+\log y) d y+y d x=0 is
View answer and explanation
Correct option: B
xy+ylog⁡y−y=cx y+y \log y-y=c
Option B (xy+ylog⁡y−y=cx y+y \log y-y=c) is correct because it is the solution obtained by solving the given differential equation.
823 Easy
823 of 975
Application +4 | -1
The integrating factor of the differential equation cos⁡x(dydx)+ysin⁡x=1\cos x\left( {\frac{{dy}}{{dx}}} \right) + y\sin x = 1 is
View answer and explanation
Correct option: A
sec⁡x\sec x
Option A (sec⁡x\sec x) is correct because it is the integrating factor derived from the given differential equation after rewriting it in standard form.
824 Medium
824 of 975
Application +4 | -1
The solution of the differential equation xdy+ydx−1−x2y2dx=0x d y+y d x-\sqrt{1-x^2 y^2} d x=0 is
View answer and explanation
Correct option: B
xy=sin⁡(x+c)x y=\sin (x+c)
Option B (xy=sin⁡(x+c)x y=\sin (x+c)) is the correct solution, obtained by integrating the differential equation after recognizing the exact differential.
825 Easy
825 of 975
Analyzing +4 | -1
Let ff be a twice differentiable function defined on RR, such that f(0)=1,    f′(0)=2f(0) = 1,\,\,\,\,f'(0) = 2 and f′′(x)≠0f''(x) \ne 0 for all x∈R.x \in R. If ∣f(x)f′(x)f′(x)f′′(x)∣=0\left| {\begin{array}{cccccccccccccccccccc} {f(x)}&{f'(x)}\\ {f'(x)}&{f''(x)} \end{array}} \right| = 0 for all x∈Rx \in R, then the value of f(1)f(1) lies in the interval
View answer and explanation
Correct option: B
(6, 9)
Option B ((6, 9)) is correct because solving the differential equation and using the initial conditions leads to f(1) = e2, which lies in the interval (6, 9).

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826 Medium
826 of 975
Analyzing +4 | -1
Family of curves y=Ax+A3y=A x+A^3 is represented by the differential equation of degree
View answer and explanation
Correct option: A
3
Option A (3) is correct because after eliminating the arbitrary constant A, the resulting differential equation has a degree of 3.
827 Easy
827 of 975
Application +4 | -1
If y=ex(sin⁡x+cos⁡x)y=e^x(\sin x+\cos x), then the value of d2ydx2−2dydx+2y\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because substituting the given y and its derivatives into the expression results in 0.
828 Easy
828 of 975
Application +4 | -1
The family of curves y=easin⁡xy = {e^{a\sin x}}, where aa is an arbitrary constant, is represented by the differential equation
View answer and explanation
Correct option: B
ylog⁡y=tan⁡xdydxy\log y = \tan x\frac{{dy}}{{dx}}
Option C (d2ydx2−m2y=0\frac{{{d^2}y}}{{d{x^2}}} - {m^2}y = 0) is correct because the given function satisfies this second-order differential equation.
829 Easy
829 of 975
Application +4 | -1
If dy dx+3cos⁡2xy=1cos⁡2x,x∈(−π3,π3)\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{3}{\cos ^2 x} y=\frac{1}{\cos ^2 x}, x \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right), and y(π4)=43y\left(\frac{\pi}{4}\right)=\frac{4}{3}, then y(−π4)y\left(-\frac{\pi}{4}\right) equals:
View answer and explanation
Correct option: A
13+e6\frac{1}{3}+e^6
Option A (13+e6\frac{1}{3}+e^6) is correct because it is the value of y(−π4-\frac{\pi}{4}) obtained after solving the given differential equation and applying the initial condition y(π4\frac{\pi}{4}) = 43\frac{4}{3}.
830 Easy
830 of 975
Understanding +4 | -1
If pp and qq are the order and degree of the differential equation ydydx+x3d2ydx2+xy=cos⁡xy \frac{d y}{d x}+x^3 \frac{d^2 y}{d x^2}+x y=\cos x, then
View answer and explanation
Correct option: C
p>qp > q
Option C is correct because the order (p) of the differential equation is 2 (highest derivative is d2y/dx2) and the degree (q) is 1 (power of the highest derivative), so p > q.
831 Easy
831 of 975
Application +4 | -1
The solution of the differential equation dydx=1+x+y+xy\frac{d y}{d x}=1+x+y+x y is:
View answer and explanation
Correct option: A
log⁡(1+y)=x+x22+c\log (1+y)=x+\frac{x^2}{2}+c
Option A (log⁡(1+y)=x+x22+c\log (1+y)=x+\frac{x^2}{2}+c) is correct because it is the result of separating variables and integrating the given differential equation.
832 Easy
832 of 975
Application +4 | -1
Equation of the curve through the origin satisfying dy=(sec⁡x+ytan⁡x)dx\mathrm{d} y=(\sec x+y \tan x) \mathrm{d} x is
View answer and explanation
Correct option: B
ycos⁡x=xy \cos x=x
Option B, ycos⁡x=xy \cos x=x, is the correct solution. It is obtained by solving the linear differential equation using an integrating factor and applying the initial condition that the curve passes through the origin.
833 Medium
833 of 975
Application +4 | -1
If y=y(x)y=y(x) is the solution of the differential equation dydx+2y=sin⁡(2x),y(0)=34,\frac{d y}{d x}+2 y=\sin (2 x), y(0)=\frac{3}{4}, then y(π8)y\left(\frac{\pi}{8}\right) is equal to:
View answer and explanation
Correct option: B
e−π/4{e}^{-\pi / 4}
The correct answer is B (e−π/4{e}^{-\pi / 4}) because it is the value of y(π/8) obtained by solving the given differential equation with the initial condition y(0) = 3/4.
834 Easy
834 of 975
Application +4 | -1
The solution of the differential equation dydx+3x21+x3y=sin⁡2x1+x3\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{\sin ^2 x}{1+x^3} is
View answer and explanation
Correct option: D
y(1+x3)=x2−14sin⁡2x+cy\left(1+x^3\right)=\frac{x}{2}-\frac{1}{4} \sin 2 x+c
Option D (y(1+x3)=x2−14sin⁡2x+cy\left(1+x^3\right)=\frac{x}{2}-\frac{1}{4} \sin 2 x+c) is the correct solution, obtained by solving the linear differential equation using an integrating factor and trigonometric identities.
835 Easy
835 of 975
Application +4 | -1
The solution of y′+cos⁡(x+y)=cos⁡(x−y)y^{\prime}+\cos (x+y)=\cos (x-y) and y(π4)=π4y\left(\frac{\pi}{4}\right)=\frac{\pi}{4} is given by
View answer and explanation
Correct option: B
cos⁡y=2sin⁡xsin⁡2y\cos y=2 \sin x \sin ^2 y
Option B (cos⁡y=2sin⁡xsin⁡2y\cos y=2 \sin x \sin ^2 y) is correct because it is the solution to the differential equation y′+cos⁡(x+y)=cos⁡(x−y)y^{\prime}+\cos (x+y)=\cos (x-y) with the initial condition y(π4)=π4y(\frac{\pi}{4})=\frac{\pi}{4}.
836 Medium
836 of 975
Application +4 | -1
Solution of the differential equation dydx=y(x−yln⁡y)x(xln⁡x−y)\frac{d y}{d x}=\frac{y(x-y \ln y)}{x(x \ln x-y)} is
View answer and explanation
Correct option: A
xln⁡x+yln⁡yxy=c\frac{x \ln x+y \ln y}{x y}=c
Option A (xln⁡x+yln⁡yxy=c\frac{{x \ln x+y \ln y}}{{x y}}=c) is correct because it is the solution obtained after integrating the given differential equation after rearrangement and simplification.
837 Easy
837 of 975
Application +4 | -1
Let y(x)y(x) be a solution of (2+sin⁡x)(1+y)dydx=cos⁡x\frac{(2+\sin x)}{(1+y)} \frac{d y}{d x}=\cos x. If y(0)=2y(0)=2, then y(π2)y\left(\frac{\pi}{2}\right) equals
View answer and explanation
Correct option: C
72\frac{7}{2}
Option C (72\frac{7}{2}) is correct because it is the value of y(π2\frac{\pi}{2}) obtained after solving the given differential equation and applying the initial condition y(0) = 2.
838 Medium
838 of 975
Application +4 | -1
If (2+sin⁡ x)dydx+(y+1)cos⁡x=0\left( {2 + \sin \,x} \right)\frac{{dy}}{{dx}} + \left( {y + 1} \right)\cos x = 0 and y(0)=1,then  y(π2)y\left( 0 \right) = 1,\mathrm{ then}\,\,y\left( {\frac{\pi }{2}} \right) is equal to
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (13\frac{1}{3}) is correct because it is the value of y(π2)y(\frac{\pi}{2}) calculated by solving the given differential equation with the initial condition.
839 Hard
839 of 975
Analyzing +4 | -1
Given dydx=yx(1+xy2(1+ln⁡x))\frac{d y}{d x}=\frac{y}{x}\left(1+x y^{2}(1+\ln x)\right). If y(1)=3y(1)=3, then the value of y2(3)9\frac{y^{2}(3)}{9} is:
View answer and explanation
Correct option: B
143+27ln⁡3\frac{1}{43+27 \ln 3}
Option A (\−143+27ln⁡3-\frac{1}{43+27 \ln 3}) is correct because it is the value of \frac{{y2(3)}}{{9}} calculated by solving the given differential equation with the initial condition y(1)=3.
840 Medium
840 of 975
Understanding +4 | -1
The differential equation of all parabolas whose axes are parallel to yy-axis, is
View answer and explanation
Correct option: A
d3ydx3=0\frac{{{d^3}y}}{{d{x^3}}} = 0
Option A (d3ydx3=0\frac{{{d^3}y}}{{d{x^3}}} = 0) is correct because the general equation of a parabola with its axis parallel to the y-axis is y=Ax2+Bx+Cy = Ax^2 + Bx + C, and its third derivative is zero.
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