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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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241 Medium
241 of 975
Application +4 | -1
The solution of the differential equation dydx=e3x−2y+x2e−2y\frac{{dy}}{{dx}} = {e^{3x - 2y}} + {x^2}{e^{ - 2y}}, is
View answer and explanation
Correct option: B
12e2y=13(e3x+x3)+c\frac{1}{2}{e^{2y}} = \frac{1}{3}\left( {{e^{3x}} + {x^3}} \right) + c
Option B (12e2y=13(e3x+x3)+c\frac{1}{2}{e^{2y}} = \frac{1}{3}\left( {{e^{3x}} + {x^3}} \right) + c) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
242 Easy
242 of 975
Application +4 | -1
The differential equation dydx=x+y−1x+y+1\frac{d y}{d x}=\frac{x+y-1}{x+y+1} reduces to variable separable form by making the substitution
View answer and explanation
Correct option: A
x+y=vx+y=v
Option A (x+y=vx+y=v) is correct because it transforms the given differential equation into a variable separable form.
243 Easy
243 of 975
Analyzing +4 | -1
A particle starts at the origin and moves along the xx-axis in such a way that its velocity at the point (x,0)\left( {x,0} \right) is given by the formula dxdt=cos⁡2πx\frac{{dx}}{{dt}} = {\cos ^2}\pi x. Then, the particle never reaches the point on
View answer and explanation
Correct option: C
x=12x = \frac{1}{2}
Option C (x=12x = \frac{1}{2}) is correct because the velocity becomes zero at this point, implying the particle stops and never reaches it.
244 Medium
244 of 975
Application +4 | -1
If dydx=x+y−2x−y\frac{d y}{d x}=\frac{x+y-2}{x-y} and y(0)=2y(0)=2, find y(2)y(2)
View answer and explanation
Correct option: A
0
Option A (0) is correct because solving the differential equation with the given initial condition leads to the solution where y(2) = 0.
245 Easy
245 of 975
Application +4 | -1
If xsin⁡y=sin⁡(y+a)x \sin y=\sin (y+a) and dydx=A1+x2−2xcos⁡a\frac{d y}{d x}=\frac{A}{1+x^2-2 x \cos a} then the value of AA is
View answer and explanation
Correct option: C
sin⁡a\sin a
Option C (sin⁡a\sin a) is correct because it is the value of A that satisfies the given relation after substituting the derivative of the function.
246 Medium
246 of 975
Application +4 | -1
Solution of the differential equation 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: A
y=keTan⁡−1(y/x)y=k e^{\operatorname{Tan}^{-1}(y / x)}
Option B (y=kxeTan⁡−1(y/x)y=k x e^{\operatorname{Tan}^{-1}(y / x)}) is incorrect. Option A (y=keTan⁡−1(y/x)y=k e^{\operatorname{Tan}^{-1}(y / x)}) is correct because it is the solution obtained after solving the homogeneous differential equation using the substitution y=vxy=vx.
247 Medium
247 of 975
Application +4 | -1
The solution of the differential equation x2dydx=x2+xy+y2x^2 \frac{d y}{d x}=x^2+x y+y^2 is
View answer and explanation
Correct option: A
tan⁡−1(yx)=log⁡x+c\tan ^{-1}\left(\frac{y}{x}\right)=\log x+c
Option A (tan⁡−1(yx)=log⁡x+c\tan ^{-1}\left(\frac{y}{x}\right)=\log x+c) is correct because it is the solution obtained after solving the homogeneous differential equation using the substitution y=vxy=vx.
248 Easy
248 of 975
Application +4 | -1
The differential equation whose solution is (x−h)2+(y−k)2=a2(x-h)^2+(y-k)^2=a^2 is ( aa is constant)
View answer and explanation
Correct option: B
[1+(y′)2]3=a2(y′′)2\left[1+\left(y^{\prime}\right)^2\right]^3=a^2\left(y^{\prime \prime}\right)^2
Option B ([1+(y′)2]3=a2(y′′)2\left[1+\left(y^{\prime}\right)^2\right]^3=a^2\left(y^{\prime \prime}\right)^2) is correct because it represents the differential equation obtained by eliminating the arbitrary constants from the given equation of a circle.
249 Easy
249 of 975
Application +4 | -1
The solution of the differential equation dydx=sin⁡(x+y)tan⁡(x+y)−1\frac{{dy}}{{dx}} = \sin \left( {x + y} \right)\tan \left( {x + y} \right) - 1 is
View answer and explanation
Correct option: B
x+cosec(x+y)=cx + \mathrm{cosec}\left( {x + y} \right) = c
Option B (x+cosec(x+y)=cx + \mathrm{cosec}\left( {x + y} \right) = c) is correct because it is the solution obtained by substituting z=x+yz = x + y and integrating the resulting equation.
250 Medium
250 of 975
Application +4 | -1
The general solution of the differential equation dydx=(1+y2)xy(1+x2)\frac{{dy}}{{dx}} = \frac{{\left( {1 + {y^2}} \right)}}{{xy\left( {1 + {x^2}} \right)}} is
View answer and explanation
Correct option: B
(1+x2)(1+y2)=cx2\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right) = c{x^2}
The correct answer is (1+x2)(1+y2)=cx2\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right) = c{x^2} because it is the result of separating variables and integrating the given differential equation.
251 Easy
251 of 975
Application +4 | -1
If y=y(x)y=y(x) is the solution of the differential equation (1+e2x)dydx+2(1+y2)ex=0\left(1+e^{2 x}\right) \frac{d y}{d x}+2\left(1+y^2\right) e^x=0 and y(0)=0y(0)=0, then 6(y′(0)+(y(log⁡e3))2)6\left(y^{\prime}(0)+\left(y\left(\log _e \sqrt{3}\right)\right)^2\right) is equal to
View answer and explanation
Correct option: C
−4-4
Option C (-4) is correct because by solving the differential equation and substituting the given values, the expression evaluates to -4.
252 Medium
252 of 975
Application +4 | -1
If dydx=xyx2+y2,y(1)=1\frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}},y\left( 1 \right) = 1, then one of the values of x0{x_0} satisfying y(x0)=ey\left( {{x_0}} \right) = e is given by
View answer and explanation
Correct option: B
e3e\sqrt 3
Option B (e3e\sqrt 3) is the correct answer because it satisfies the given differential equation and initial condition when y(x0)=ey(x_0) = e.
253 Medium
253 of 975
Application +4 | -1
The solution of the differential equation xlog⁡xdydx+y=2log⁡xx \log x \frac{d y}{d x}+y=2 \log x is
View answer and explanation
Correct option: C
ylog⁡x=(log⁡x)2+cy \log x=(\log x)^2+c
Option C (ylog⁡x=(log⁡x)2+cy \log x=(\log x)^2+c) is correct because it represents the solution to the given differential equation after finding the integrating factor and integrating.
254 Easy
254 of 975
Application +4 | -1
The differential equation of the family of non-vertical lines passing through the origin is
View answer and explanation
Correct option: B
xdydx−y=0x \frac{d y}{d x}-y=0
Option B (xdydx−y=0x \frac{{dy}}{{dx}} - y = 0) is correct because it represents the differential equation of all non-vertical lines passing through the origin.
255 Medium
255 of 975
Application +4 | -1
The solution of the differential equation (x+y)(dx−dy)=dx+dy(x+y)(d x-d y)=d x+d y, is
View answer and explanation
Correct option: D
x+y=kex−yx+y=k e^{x-y}
Option D (x+y=kex−yx+y=k e^{x-y}) is the correct solution obtained by separating variables and integrating the given differential equation.

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256 Easy
256 of 975
Application +4 | -1
Solution of y(dy dx)2+2xdy dx=yy\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2+2 x \frac{\mathrm{d} y}{\mathrm{~d} x}=y is
View answer and explanation
Correct option: C
y2=2Cx+C2y^2=2 C x+C^2
Option C (y2=2Cx+C2y^2=2 C x+C^2) is correct because it is the solution obtained by solving the given differential equation using Clairaut's form.
257 Medium
257 of 975
Application +4 | -1
The differential equation dydx=1−y2y\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} determine of circles with
View answer and explanation
Correct option: C
fixed radius 1 and variable centres along the xx-axis
Option C (fixed radius 1 and variable centres along the xx-axis) is correct because the solution to the differential equation represents circles with a radius of 1 and centers on the x-axis.
258 Medium
258 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (x−x3)dy=(y+yx2−3x4)dx,x>2\left(x-x^3\right) d y=\left(y+y x^2-3 x^4\right) d x, x>2. If y(3)=3y(3)=3, then y(4)y(4) is equal to
View answer and explanation
Correct option: C
12
Option C (12) is correct because by solving the given differential equation and using the initial condition y(3)=3, we find y(4) = 12.
259 Medium
259 of 975
Application +4 | -1
y=aemx+be−mxy = a{e^{mx}} + b{e^{ - mx}} satisfies which of the following differential equations
View answer and explanation
Correct option: C
d2ydx2−m2y=0\frac{{{d^2}y}}{{d{x^2}}} - {m^2}y = 0
Option C (d2ydx2−m2y=0\frac{{{d^2}y}}{{d{x^2}}} - {m^2}y = 0) is correct because it is the differential equation that the given function satisfies after taking the second derivative.
260 Easy
260 of 975
Application +4 | -1
If y(t)y\left( t \right) is a solution of (1+t)dydt−ty=1\left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1 and y(0)=−1y\left( 0 \right) = - 1, then y(1)y\left( 1 \right) is equal to
View answer and explanation
Correct option: A
−12- \frac{1}{2}
Option A (−12- \frac{1}{2}) is correct because it is the value of y(1) obtained by solving the given differential equation with the initial condition y(0) = -1.
261 Medium
261 of 975
Application +4 | -1
The solution of the differential equation dydx=ey+x+ey−x\frac{{dy}}{{dx}} = {e^{y + x}} + {e^{y - x}} is
View answer and explanation
Correct option: B
e−y=e−x−ex+c{e^{ - y}} = {e^{ - x}} - {e^x} + c
Option B (e−y=e−x−ex+c{e^{ - y}} = {e^{ - x}} - {e^x} + c) is correct because it represents the solution of the given differential equation after separation of variables and integration.
262 Easy
262 of 975
Application +4 | -1
Integrating factor of the differential equation dydx+ytan⁡x−sec⁡x=0\frac{d y}{d x}+y \tan x-\sec x=0 is
View answer and explanation
Correct option: C
1cos⁡x\frac{1}{\cos x}
Option C (1cos⁡x\frac{{1}}{{\cos x}}) is correct because it is the integrating factor of the given linear differential equation after rearranging it.
263 Medium
263 of 975
Application +4 | -1
The solution of the equation dydx=x+yx−y\frac{d y}{d x}=\frac{x+y}{x-y} is
View answer and explanation
Correct option: B
c(x2+y2)1/2=etan⁡−1(y/x)c\left(x^2+y^2\right)^{1 / 2}=e^{\tan ^{-1}(y / x)}
Option B (c(x2+y2)1/2=etan⁡−1(y/x)c(x^2+y^2)^{1/2}=e^{\tan^{-1}(y/x)}) is the correct solution, obtained by solving the homogeneous differential equation using the substitution y=vxy=vx.
264 Easy
264 of 975
Application +4 | -1
If y=y(x)y=y(x) is the solution of the differential equation (1+e2x)dydx+2(1+y2)ex=0\left(1+e^{2 x}\right) \frac{d y}{d x}+2\left(1+y^2\right) e^x=0 and y(0)=0y(0)=0 then which of the following is correct
View answer and explanation
Correct option: A
2tan⁡−1(ex)+tan⁡−1y=π22 \tan ^{-1}\left(e^x\right)+\tan ^{-1} y=\frac{\pi}{2}
Option A, 2tan⁡−1(ex)+tan⁡−1y=π22 \tan ^{-1}\left(e^x\right)+\tan ^{-1} y=\frac{\pi}{2}, is the correct solution obtained by separating variables, integrating, and applying the initial condition.
265 Easy
265 of 975
Analyzing +4 | -1
If y(x−y)2=xy(x-y)^2=x, then
∫dx(x−3y)=mnln⁡[(x−y)2−1]\int \frac{d x}{(x-3 y)}=\frac{m}{n} \ln \left[(x-y)^2-1\right].Then (m+2n)(m+2 n) is equal to
View answer and explanation
Correct option: C
5
Option C (5) is correct because, after differentiating the given equation and substituting into the integral, the value of m+2n is found to be 5.
266 Medium
266 of 975
Application +4 | -1
The solution of y′=1+x+y2+xy2,y(0)=0y' = 1 + x + {y^2} + x{y^2},y\left( 0 \right) = 0 is
View answer and explanation
Correct option: D
y=tan⁡(x+x22)y = \tan \left( {x + \frac{{{x^2}}}{2}} \right)
Option D, y=tan⁡(x+x22)y = \tan \left( {x + \frac{{{x^2}}}{2}} \right), is the correct solution because it satisfies the given differential equation and the initial condition y(0)=0y(0) = 0.
267 Medium
267 of 975
Application +4 | -1
y=2e2x−e−xy = 2{e^{2x}} - {e^{ - x}} is a solution of the differential equation
View answer and explanation
Correct option: D
y2−y1−2y=0{y_2} - {y_1} - 2y = 0
Option D (y2−y1−2y=0{y_2} - {y_1} - 2y = 0) is correct because it is the differential equation that is satisfied by the given function y=2e2x−e−xy = 2{e^{2x}} - {e^{ - x}}.
268 Easy
268 of 975
Application +4 | -1
The population p(t)p(t) at time tt of a certain mouse species follows the differential equation dp(t)dt=0.5\frac{d p(t)}{d t}=0.5 p(t)−450p(t)-450. If p(0)=850\mathrm{p}(0)=850, then the time at which the population becomes zero is
View answer and explanation
Correct option: D
2log⁡182 \log 18
Option D (2log⁡182 \log 18) is correct because it represents the time at which the population becomes zero, obtained by solving the differential equation and applying the given initial condition.
269 Medium
269 of 975
Analyzing +4 | -1
The solution of the differential equation y1y3=3y22{y_1}{y_3} = 3y_2^2 is
View answer and explanation
Correct option: A
x=A1y2+A2y+A3x = {A_1}{y^2} + {A_2}y + {A_3}
Option A (x=A1y2+A2y+A3x = {A_1}{y^2} + {A_2}y + {A_3}) is correct because it represents the general solution obtained by integrating the given differential equation twice.
270 Easy
270 of 975
Understanding +4 | -1
Which of the following equation is non-linear
View answer and explanation
Correct option: D
xdydx+3dydx=y2x \frac{d y}{d x}+\frac{3}{\frac{d y}{d x}}=y^2
Option D (xdydx+3dydx=y2x \frac{{d y}}{{d x}}+\frac{{3}}{{\frac{{d y}}{{d x}}}}=y^2) is non-linear because it contains a term with the derivative in the denominator, which makes it non-linear.
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