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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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601 Hard
601 of 975
Application +4 | -1
Let a differentiable function ff satisfy f(x)+∫3xf(t)tdt=x+1,x≥3f(x)+\int\limits_{{3}}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3. Then 12f(8)12 f(8) is equal to
View answer and explanation
Correct option: C
17
The correct answer is C (17) because it is the value of 12f(8) obtained by solving the given differential equation.
602 Hard
602 of 975
Application +4 | -1
The general solution of the differential equation d2ydx2+2dydx+y=2e3x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} + y = 2{e^{3x}} is given by
View answer and explanation
Correct option: C
y=(c1+c2x)e−x+e3x8y = \left( {{c_1} + {c_2}x} \right){e^{ - x}} + \frac{{{e^{3x}}}}{8}
Option C (y=(c1+c2x)e−x+e3x8y = \left( {{c_1} + {c_2}x} \right){e^{ - x}} + \frac{{{e^{3x}}}}{8}) is correct because it represents the general solution of the given second-order linear homogeneous differential equation with constant coefficients.
603 Medium
603 of 975
Application +4 | -1
The solution of the equation x2d2ydx2=log⁡x{x^2}\frac{{{d^2}y}}{{d{x^2}}} = \log x when x=1,y=0x = 1,y = 0 and dydx=−1\frac{{dy}}{{dx}} = - 1 is
View answer and explanation
Correct option: D
−12(log⁡x)2−log⁡x- \frac{1}{2}{\left( {\log x} \right)^2} - \log x
Option D (−12(log⁡x)2−log⁡x- \frac{1}{2}{\left( {\log x} \right)^2} - \log x) is correct because it is the solution to the given second-order differential equation with the given initial conditions.
604 Medium
604 of 975
Application +4 | -1
The solution of the differential equation cos⁡xdy=y(sin⁡x−y)dx\cos x d y=y(\sin x-y) d x is ...... .
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Correct option: D
sec⁡x=(tan⁡x+c)y\sec x=(\tan x+c) y
Option D (sec⁡x=(tan⁡x+c)y\sec x=(\tan x+c) y) is the correct solution, found by recognizing the equation as a Bernoulli equation and using the appropriate substitution.
605 Medium
605 of 975
Application +4 | -1
The solution of differential equation yexydx=(xexy+y2)dyy e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^2\right) d y is
View answer and explanation
Correct option: A
y=exy+Cy=e^{\frac{x}{y}}+C
Option A (y=exy+Cy=e^{\frac{x}{y}}+C) is the correct solution obtained by solving the given homogeneous differential equation using substitution and integration.
606 Medium
606 of 975
Application +4 | -1
Solution of x(dy/dx)−yx2−y2=mx2\frac{x(d y / d x)-y}{\sqrt{x^2-y^2}}=m x^2 is
View answer and explanation
Correct option: B
Sin⁡−1(yx)=mx22+c\operatorname{Sin}^{-1}\left(\frac{y}{x}\right)=\frac{m x^2}{2}+c
Option B (Sin⁡−1(yx)=mx22+c\operatorname{Sin}^{-1}(\frac{y}{x})=\frac{m x^2}{2}+c) is correct because it is the solution obtained after solving the differential equation using the substitution y=vxy=vx.
607 Easy
607 of 975
Application +4 | -1
The solution of dydx=ax+bcy+d\frac{d y}{d x}=\frac{a x+b}{c y+d} represents a parabola if
View answer and explanation
Correct option: C
a=0,c≠0a=0, c \neq 0
Option C, a=0,c≠0a=0, c \neq 0, is correct because it results in an equation of the form cy2+dy=bx+Kcy^2 + dy = bx + K, which represents a parabola.
608 Easy
608 of 975
Application +4 | -1
The solution of x dyx2+y2=(yx2+y2−1)\frac{x \mathrm{~d} y}{x^2+y^2}=\left(\frac{y}{x^2+y^2}-1\right) dx\mathrm{d} x is
View answer and explanation
Correct option: C
y=xtan⁡(c−x)y=x \tan (c-x)
Option C (y=xtan⁡(c−x)y=x \tan (c-x)) is correct because it is the solution obtained by rewriting the given differential equation, separating variables, and integrating.
609 Easy
609 of 975
Application +4 | -1
The solution of the differential equation dydx=ex+y\frac{{dy}}{{dx}} = {e^{x + y}} is
View answer and explanation
Correct option: C
ex+e−y=c{e^x} + {e^{ - y}} = c
Option C (ex+e−y=c{e^x} + {e^{ - y}} = c) is correct because it is the solution to the given separable differential equation.
610 Medium
610 of 975
Application +4 | -1
The order and the degree of the differential equation of the family of parabolas whose major axis is X\mathrm{X}-axis are and respectively .
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Correct option: D
2,1
Option D (2,1) is correct because the order of the differential equation is 2 (second derivative) and the degree is 1 (highest power of the highest derivative).
611 Easy
611 of 975
Application +4 | -1
The value of ∫0πsin⁡3xcos⁡2xdx\int_0^\pi \sin ^3 x \cos ^2 x d x is
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Correct option: B
415\frac{4}{15}
Option B (415\frac{4}{15}) is correct because it uses the properties of trigonometric integrals and Wallis' formula to efficiently compute the definite integral.
612 Medium
612 of 975
Application +4 | -1
The solution of the equation dy dx=\frac{\mathrm{d} y}{\mathrm{~d} x}= cos⁡(x−y)\cos (x-y) is
View answer and explanation
Correct option: B
x+cot⁡(x−y2)=cx+\cot \left(\frac{x-y}{2}\right)=c
Option C (x+tan⁡(x−y2)=cx+\tan \left(\frac{x-y}{2}\right)=c) is incorrect. Option D (none of these) is correct because the correct solution after integration and simplification is x+cot⁡(x−y2)=Cx + \cot(\frac{x-y}{2}) = C.
613 Easy
613 of 975
Application +4 | -1
If xpyq=(x+y)p+qx^p y^q=(x+y)^{p+q}, then dy/dxd y / d x is equal to
View answer and explanation
Correct option: A
yx\frac{y}{x}
Option A (yx\frac{y}{x}) is correct because by taking the logarithm of both sides and differentiating implicitly, we find that dydx=yx\frac{dy}{dx} = \frac{y}{x}.
614 Easy
614 of 975
Application +4 | -1
The solution curve of the differential equation, (1+e−x)(1+y2)dydx=y2(1 + {e^{ - x}})(1 + {y^2})\frac{{dy}}{{dx}} = {y^2}, which passes through the point (0, 1) is
View answer and explanation
Correct option: C
y2=1+ylog⁡e[1+ex2]{y^2} = 1 + y{\log _e}\left[ {\frac{{1 + {e^x}}}{2}} \right]
Option C (y2=1+ylog⁡e[1+ex2]{y^2} = 1 + y{\log _e}\left[ {\frac{{1 + {e^x}}}{2}} \right]) is correct because it represents the solution to the given differential equation that passes through the point (0, 1).
615 Easy
615 of 975
Application +4 | -1
The general solution of the differential equation (1+y2)dx+(1+x2)dy=0\left( {1 + {y^2}} \right)dx + \left( {1 + {x^2}} \right)dy = 0 is
View answer and explanation
Correct option: C
x+y=c(1−xy)x + y = c\left( {1 - xy} \right)
Option D (x+y=c(1−xy)x + y = c(1 - xy)) is correct because it is the general solution obtained by separating variables and integrating the given differential equation.

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616 Easy
616 of 975
Application +4 | -1
Let us consider a curve, y=f(x)y = f(x) passing through the point (−2,  2)( - 2,\,\,2) and the slope of the tangent to the curve at any point (x,f(x))(x,f(x)) is given by f(x)+xf′(x) =x2f(x) + xf'(x)\, = {x^2} then
View answer and explanation
Correct option: C
x3−3xf(x)−4=0{x^3} - 3x\mathrm{ }f(x) - 4 = 0
The correct answer is x3−3xf(x)−4=0{x^3} - 3x\mathrm{ }f(x) - 4 = 0 because it is obtained by solving the given differential equation and using the given point to find the constant of integration.
617 Medium
617 of 975
Application +4 | -1
Tangent at any point P(x,y)P(x, y) of a curve y=f(x)y=f(x) meets the yy-axis in BB. If the algebraic value of OBO B is equal to x+2x+2, then the equation of the curve is
View answer and explanation
Correct option: A
y+2=cx−xlog⁡∣x∣y+2=c x-x \log |x|
Option A (y+2=cx−xlog⁡∣x∣y+2=c x-x \log |x|) is correct because it represents the solution to the differential equation derived from the given geometric condition, solved using an integrating factor.
618 Easy
618 of 975
Application +4 | -1
A differential equation associated with the primitive y=a+be5x+ce−7xy=a+b e^{5 x}+c e^{-7 x} is
View answer and explanation
Correct option: C
y3+2y2−35y1=0y_3+2 y_2-35 y_1=0
Option C (y3+2y2−35y1=0y_3+2 y_2-35 y_1=0) is correct because it is the differential equation associated with the given primitive.
619 Easy
619 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 (-12\frac{{1}}{{2}}) is correct because, after solving the differential equation and applying the initial condition, y(1) equals -12\frac{{1}}{{2}}.
620 Easy
620 of 975
Application +4 | -1
If the integrating factor of the differential equation dydx+P(x)y=Q(x)\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) is xx, then P(x)P\left( x \right) is
View answer and explanation
Correct option: C
1x\frac{1}{x}
Option C (1x\frac{1}{x}) is correct because the integrating factor is given as x, and P(x) can be found by differentiating the logarithm of the integrating factor.
621 Easy
621 of 975
Application +4 | -1
The solution of the differential equation dydx+1=ex+y\frac{d y}{d x}+1=e^{x+y}, is
View answer and explanation
Correct option: D
(x−C)ex+y+1=0(x-C) e^{x+y}+1=0
Option D ((x−C)ex+y+1=0(x-C) e^{x+y}+1=0) is correct because it is the solution obtained by solving the given differential equation using a suitable substitution.
622 Hard
622 of 975
Application +4 | -1
The solution of the differential equation 2yexydx−(y+2xeyy)dy=02 y e^{\frac{x}{y}} d x-\left(y+2 x e^{\frac{y}{y}}\right) d y=0 is
View answer and explanation
Correct option: A
2cexy=log⁡y2 c e^{\frac{x}{y}}=\log y
Option A (2cexy=log⁡y2 c e^{\frac{x}{y}}=\log y) is the correct solution, found by recognizing the equation as homogeneous and using the appropriate substitution.
623 Medium
623 of 975
Application +4 | -1
The solution of the differential equation xdx+ydyxdx−ydy=y3x3\frac{\sqrt{x} d x+\sqrt{y} d y}{\sqrt{x} d x-\sqrt{y} d y}=\sqrt{\frac{y^3}{x^3}} is given by,
View answer and explanation
Correct option: D
12log⁡(x3+y3)+tan⁡−1(yx)3/2=c\frac{1}{2} \log \left(x^3+y^3\right)+\tan ^{-1}\left(\frac{y}{x}\right)^{3 / 2}=c
Option D (12log⁡(x3+y3)+tan⁡−1(yx)3/2=c\frac{{1}}{{2}} \log \left(x^3+y^3\right)+\tan ^{{-1}}\left(\frac{{y}}{{x}}\right)^{{3 / 2}}=c) is correct because it represents the solution to the given differential equation, obtained through homogeneous substitution and integration.
624 Easy
624 of 975
Application +4 | -1
If the solution of differential equation x2d2ydx2+2xdydx=12yx^2 \frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}=12 y is y=Axm+Bx−ny=A x^m+B x^{-n} then find the value of m+nm+n, if m & n∈Nm \,\&\, n \in N.
View answer and explanation
Correct option: A
7
Option A (7) is correct because by substituting the given solution into the differential equation and solving for m and n, we find that m+n = 7.
625 Medium
625 of 975
Application +4 | -1
The general solution of the equation (ey+1)cos⁡xdx+eysin⁡xdy=0\left(e^y+1\right) \cos x d x+e^y \sin x d y=0 is:
View answer and explanation
Correct option: C
(ey+1)sin⁡x=c\left(e^y+1\right) \sin x=c
Option C ((ey+1e^y+1) sin⁡x=c\sin x=c) is correct because it is the general solution obtained by separating variables and integrating the given differential equation.
626 Easy
626 of 975
Application +4 | -1
The solution of the equation (x+y)dy−(x−y)(x+y) \mathrm{d} y-(x-y) dx=0\mathrm{d} x=0 is
View answer and explanation
Correct option: C
x3+2xy−y2=Kx^3+2 x y-y^2=K
The correct answer is not listed in the options. The closest option is B (y2+2xy−x2=Ky^2+2 x y-x^2=K), but it is incorrect. The correct solution is x2−2xy−y2=Kx^2 - 2xy - y^2 = K.
627 Medium
627 of 975
Application +4 | -1
The solution of differential equation y−xdydx=a(y2+dydx)y - x\frac{{dy}}{{dx}} = a\left( {{y^2} + \frac{{dy}}{{dx}}} \right) is
View answer and explanation
Correct option: B
(x+a)(1−ay)=cy\left( {x + a} \right)\left( {1 - ay} \right) = cy
Option B ((x+a)(1−ay)=cy\left( {x + a} \right)\left( {1 - ay} \right) = cy) is correct because it is the solution obtained after separating variables and integrating the given differential equation.
628 Medium
628 of 975
Application +4 | -1
If dydx=1+x+y+xy\frac{{dy}}{{dx}} = 1 + x + y + xy and y(−1)=0y\left( { - 1} \right) = 0, then the function yy is
View answer and explanation
Correct option: B
e(1+x)2/2−1{e^{{{\left( {1 + x} \right)}^2}/2}} - 1
Option B (e(1+x)2/2−1{e^{{{\left( {1 + x} \right)}^2}/2}} - 1) is correct because it satisfies the given differential equation and the initial condition y(-1) = 0.
629 Medium
629 of 975
Application +4 | -1
The general solution of the equation dydx+yx=x2\frac{d y}{d x}+\frac{y}{x}=x^2 is
View answer and explanation
Correct option: C
xy=x44+cx y=\frac{x^4}{4}+c
Option C (xy=x44+cx y=\frac{x^4}{4}+c) is correct because it is the solution obtained after solving the given linear differential equation.
630 Medium
630 of 975
Application +4 | -1
The general solution of 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: D
log⁡y=tan⁡−1yx+C\log y=\tan ^{-1} \frac{y}{x}+C
The correct answer is D (log⁡y=tan⁡−1yx+C\log y=\tan ^{-1} \frac{y}{x}+C) because it is the solution obtained by solving the given homogeneous differential equation using the substitution y=vxy=vx.
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