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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Medium

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211 Medium
211 of 452
Analyzing +4 | -1
The differential equation of the family of parabola with focus as the origin and the axis as xx-axis, is
View answer and explanation
Correct option: B
−y(dydx)2=2xdydx−y- y{\left( {\frac{{dy}}{{dx}}} \right)^2} = 2x\frac{{dy}}{{dx}} - y
Option B (−y(dydx)2=2xdydx−y- y{\left( {\frac{{dy}}{{dx}}} \right)^2} = 2x\frac{{dy}}{{dx}} - y) is the correct differential equation for the family of parabolas with focus at the origin and axis as the x-axis.
212 Medium
212 of 452
Analyzing +4 | -1
The differential equation obtained by eliminating the arbitrary constants aa and bb from xy=aex+be−xxy = a{e^x} + b{e^{ - x}} is
View answer and explanation
Correct option: A
xd2ydx2+2dydx−xy=0x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy = 0
Option A (xd2ydx2+2dydx−xy=0x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy = 0) is correct because it represents the differential equation obtained after differentiating the given equation twice and eliminating the arbitrary constants a and b.
213 Medium
213 of 452
Application +4 | -1
The solution of the differential equation xdy−ydx=x3+y2dxx\mathrm{ }dy - y\mathrm{ }dx = \sqrt {{x^3} + {y^2}} dx, is
View answer and explanation
Correct option: D
None of these
Option B (xx) is correct because it is the integrating factor of the given differential equation after rearranging it into the standard form.
214 Medium
214 of 452
Understanding +4 | -1
The differential equation of all non-vertical lines in a plane is
View answer and explanation
Correct option: A
d2ydx2=0\frac{{{d^2}y}}{{d{x^2}}} = 0
Option A (d2ydx2=0\frac{{{d^2}y}}{{d{x^2}}} = 0) is correct because the general equation of a non-vertical line is ax+by=1ax + by = 1, and its second derivative with respect to x is zero.
215 Medium
215 of 452
Application +4 | -1
The order of differential equation of all parabola with it’s axis parallel to yy-axis and touch xx-axis is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the general equation of such a parabola contains two arbitrary constants.
216 Medium
216 of 452
Application +4 | -1
The solution of the equation dydx=yx(log⁡yx+1)\frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right) is
View answer and explanation
Correct option: A
log⁡(yx)=cx\log \left(\frac{y}{x}\right)=c x
Option A (log⁡(yx)=cx\log \left(\frac{y}{x}\right)=c x) is the correct solution to the differential equation, obtained by substituting y=vxy=vx and integrating.
217 Medium
217 of 452
Application +4 | -1
A solution of y=2x(dy dx)+x2(dy dx)4y=2 x\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)+x^2\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^4 is
View answer and explanation
Correct option: D
y=2cx+c2y=2 \sqrt{c x}+c^2
The correct answer is D (y=2cx+c2y=2 \sqrt{{c x}}+c^2) because it satisfies the given differential equation when p=cxp = \sqrt{{\frac{{c}}{{x}}}} is substituted.
218 Medium
218 of 452
Understanding +4 | -1
Observe the following statements

A:Interating factor of dydx+y=x2\frac{{dy}}{{dx}} + y = {x^2} is ex{e^x}
R: Integrating factor of dydx+P(x)y=Q(x)\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) is e∫P(x)dx{e^{\int P\left( x \right)dx}}
Then, the true statement among the following is
View answer and explanation
Correct option: C
A is true, R is true, R⇒A\mathrm{R} \Rightarrow \mathrm{A}
Option C (A is true, R is true, R⇒A\mathrm{R} \Rightarrow \mathrm{A}) is correct because both the integrating factor for the given equation and the general formula for the integrating factor are correct, and the general formula implies the specific case.
219 Medium
219 of 452
Application +4 | -1
The differential equation for all family of lines which are at a unit distance from the origin is:
View answer and explanation
Correct option: C
(y−xdydx)2=1+(dydx)2\left(y-x \frac{d y}{d x}\right)^2=1+\left(\frac{d y}{d x}\right)^2
Option C ((y−xdydx)2=1+(dydx)2(y-x \frac{{dy}}{{dx}})^2=1+(\frac{{dy}}{{dx}})^2) is correct because it represents the differential equation for the family of lines at a unit distance from the origin, derived by eliminating the parameter from the general equation of such lines.
220 Medium
220 of 452
Application +4 | -1
The solution of the differential equation dydx=xlog⁡x\frac{{dy}}{{dx}} = x\log x is
View answer and explanation
Correct option: B
y=x22log⁡x−x24+cy = \frac{{{x^2}}}{2}\log x - \frac{{{x^2}}}{4} + c
Option B (y=x22log⁡x−x24+cy = \frac{{{x^2}}}{2}\log x - \frac{{{x^2}}}{4} + c) is correct because it is the solution to the given differential equation obtained by integrating by parts.
221 Medium
221 of 452
Application +4 | -1
The general solution of y2dx+(x2−xy+y2)dy=0{y^2}dx + \left( {{x^2} - xy + {y^2}} \right)dy = 0 is
View answer and explanation
Correct option: A
tan⁡−1(xy)+log⁡y+c=0{\tan ^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0
Option A (tan⁡−1(xy)+log⁡y+c=0{\tan ^{ - 1}}(\frac{x}{y}) + \log y + c = 0) is correct because it is the general solution obtained by solving the homogeneous differential equation.
222 Medium
222 of 452
Application +4 | -1
The solution of y2−2y1+y=0y_2-2 y_1+y=0 is
View answer and explanation
Correct option: B
y=(c1+c2x)exy=\left(c_1+c_2 x\right) e^x
Option B (y=(c1+c2x)exy=\left(c_1+c_2 x\right) e^x) is correct because it represents the general solution to the given second-order homogeneous linear differential equation with repeated roots.
223 Medium
223 of 452
Analyzing +4 | -1
Let f(x)=(sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x))2−1,∣x∣>1f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^2-1,|x| > 1. If dydx=12ddx(sin⁡−1(f(x)))\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1}(f(x))\right) and y(3)=π6y(\sqrt{3})=\frac{\pi}{6}, then y(−3)y(-\sqrt{3}) is equal to:
View answer and explanation
Correct option: B
−π6-\frac{\pi}{6}
Option B (-π6\frac{{\pi}}{{6}}) is correct because it's the value of y(-3\sqrt{{3}}) obtained after simplification and substitution into the given equation.
224 Medium
224 of 452
Application +4 | -1
The solution of the differential equation dydx−ytan⁡x=exsec⁡x\frac{{dy}}{{dx}} - y\tan x = {e^x}\sec x is
View answer and explanation
Correct option: B
ycos⁡x=ex+cy\cos x = {e^x} + c
Option B (ycos⁡x=ex+cy\cos x = {e^x} + c) is correct because it is the solution obtained by finding the integrating factor and solving the linear differential equation.
225 Medium
225 of 452
Application +4 | -1
Integrating factor of equation (x2+1)dydx+2xy=x2−1\left(x^2+1\right) \frac{d y}{d x}+2 x y=x^2-1 is
View answer and explanation
Correct option: A
x2+1x^2+1
Option A (x2+1x^2+1) is correct because it is the integrating factor obtained by solving the differential equation.
226 Medium
226 of 452
Application +4 | -1
The solution of the differential equation 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 correct because it is the solution to the given differential equation with the initial condition y(0) = 0.
227 Medium
227 of 452
Application +4 | -1
The solution of the differential equation dydx=ex−y+x2e−y\frac{{dy}}{{dx}} = {e^{x - y}} + {x^2}{e^{ - y}} is
View answer and explanation
Correct option: B
ey−ex=13x3+c{e^y} - {e^x} = \frac{1}{3}{x^3} + c
Option B (ey−ex=13x3+c{e^y} - {e^x} = \frac{1}{3}{x^3} + c) is correct because it is the solution to the given separable differential equation.
228 Medium
228 of 452
Application +4 | -1
The solution of the equation dydx+x(x+y)=x3(x+y)3−1\frac{d y}{d x}+x(x+y)=x^3(x+y)^3-1 is
View answer and explanation
Correct option: C
1(x+y)2=x2+1+cex2\frac{1}{(x+y)^2}=x^2+1+c e^{x^2}
Option C (1(x+y)2=x2+1+cex2\frac{{1}}{{(x+y)^2}}=x^2+1+c e^{{x}^2}) is correct because it represents the solution obtained by substituting v=x+yv = x+y and solving the resulting differential equation.
229 Medium
229 of 452
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (x2−3y2)dx+3xy dy=0, y(1)=1\left(x^{2}-3 y^{2}\right) d x+3 x y \, d y=0, \, y(1)=1. Then 6y2(e)6 y^{2}(e) is equal to
View answer and explanation
Correct option: D
2e22 e^{2}
Option D (2e22e^2) is correct because it involves solving a homogeneous differential equation with an initial condition.
230 Medium
230 of 452
Application +4 | -1
The solution of y′−y=1,y(0)=−1y' - y = 1,y\left( 0 \right) = - 1 is given by y(x)y\left( x \right), which is equal to
View answer and explanation
Correct option: C
−1- 1
Option A (3) is correct because it is the speed of the object in km/h calculated from the change in object position and the given time.
231 Medium
231 of 452
Application +4 | -1
If dydx+y=2e2x\frac{{dy}}{{dx}} + y = 2{e^{2x}}, then yy is equal to
View answer and explanation
Correct option: C
ce−x+23e2xc{e^{ - x}} + \frac{2}{3}{e^{2x}}
Option C (ce−x+23e2xc{e^{ - x}} + \frac{2}{3}{e^{2x}}) is correct because it is the solution to the given first-order linear differential equation obtained using the integrating factor method.
232 Medium
232 of 452
Application +4 | -1
Let II be the purchase value of an equipment and V(t)V(t) be the value after it has been used for tt years. The value V(t)V(t) depreciates at a rate given by the differential equation dVdt=−k(T−t)\frac{d V}{d t}=-k(T-t), where k>0k>0 is a constant and TT is the total life in years of the equipment. Then the scrap value V(T)V(T) of the equipment is:
View answer and explanation
Correct option: B
I−kT22I-\frac{k T^2}{2}
Option B (I−kT22I-\frac{{k T^2}}{{2}}) is correct because it represents the scrap value V(T) obtained by solving the given differential equation with the initial condition V(0) = I.
233 Medium
233 of 452
Application +4 | -1
The differential equation of the family of curve y2=4a(x+a){y^2} = 4a\left( {x + a} \right), is
View answer and explanation
Correct option: C
y2(dydx)2+2xydydx−y2=0{y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2xy\frac{{dy}}{{dx}} - {y^2} = 0
Option C (y2(dydx)2+2xydydx−y2=0{y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2xy\frac{{dy}}{{dx}} - {y^2} = 0) is correct because it is the differential equation obtained by eliminating the parameter 'a' from the given family of curves.
234 Medium
234 of 452
Application +4 | -1
The length of the subnormal of the curve x2=4yx^2=4 y at point (−2,1)(-2,1) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the length of the subnormal is calculated using the formula y×dydxy \times \frac{{dy}}{{dx}} evaluated at the given point.
235 Medium
235 of 452
Application +4 | -1
The solution of the differential equation (1+y2)+(x−ettan⁡−1y)dydx=0\left( {1 + {y^2}} \right) + \left( {x - {e^{{\mathop\mathrm{t}\nolimits} {{\tan }^{ - 1}}y}}} \right)\frac{{dy}}{{dx}} = 0 is
View answer and explanation
Correct option: A
2xetan⁡−1y=e2tan⁡−1y+c2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + c
Option A (2xetan⁡−1y=e2tan⁡−1y+c2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + c) is correct because it is the solution obtained by solving the given linear differential equation using the integrating factor method.
236 Medium
236 of 452
Application +4 | -1
d2xdy2\frac{d^2 x}{d y^2} equals to
View answer and explanation
Correct option: B
−(d2ydx2)(dydx)−3-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}
Option B (−(d2ydx2)(dydx)−3-\left(\frac{{d^2 y}}{{d x^2}}\right)\left(\frac{{d y}}{{d x}}\right)^{-3}) is correct because it is the result of applying the chain rule and inverse function differentiation rules.
237 Medium
237 of 452
Application +4 | -1
The solution of dydx=(yx)13\frac{{dy}}{{dx}} = {\left( {\frac{y}{x}} \right)^{\frac{1}{3}}}, is
View answer and explanation
Correct option: C
y2/3−x2/3=C{y^{2/3}} - {x^{2/3}} = C
Option C (y2/3−x2/3=C{y^{2/3}} - {x^{2/3}} = C) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
238 Medium
238 of 452
Application +4 | -1
The solution of the equation dy/dx=cos⁡(x−y)d y / d x=\cos (x-y), where y(3π2)=πy\left(\frac{3 \pi}{2}\right)=\pi is
View answer and explanation
Correct option: B
x+cot⁡(x−y2)=1+3π2x+\cot \left(\frac{x-y}{2}\right)=1+\frac{3 \pi}{2}
Option B (x+cot⁡(x−y2)=1+3π2x+\cot \left(\frac{{x-y}}{{2}}\right)=1+\frac{{3 \pi}}{{2}}) is the correct solution obtained by solving the given differential equation and applying the initial condition.
239 Medium
239 of 452
Application +4 | -1
The equation of the curve whose subnormal is equal to a constant aa is
View answer and explanation
Correct option: B
y2=2ax+2b{y^2} = 2ax + 2b
Option B (y2=2ax+2b{y^2} = 2ax + 2b) is correct because it is the solution obtained by integrating the differential equation formed from the given condition about the subnormal.
240 Medium
240 of 452
Application +4 | -1
The solution of the D.E ydx−xdy+3x2y2dx=0y d x-x d y+3 x^2 y^2 d x=0 is
View answer and explanation
Correct option: C
x3y+x−cy=0x^3 y+x-c y=0
Option C (x3y+x−cy=0x^3 y+x-c y=0) is correct because it represents the solution obtained by integrating the given differential equation after rewriting it in a suitable form.
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