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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Medium

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271 Medium
271 of 452
Application +4 | -1
Solution of dydx=xlog⁡x2+xsin⁡y+ycos⁡y\frac{{dy}}{{dx}} = \frac{{x\log {x^2} + x}}{{\sin y + y\cos y}} is
View answer and explanation
Correct option: A
ysin⁡y=x2log⁡x+cy\sin y = {x^2}\log x + c
Option A (ysin⁡y=x2log⁡x+cy\sin y = {x^2}\log x + c) is correct because it represents the solution obtained by separating variables and integrating the given differential equation.
272 Medium
272 of 452
Application +4 | -1
The differential equation ydydx+x=cy\frac{{dy}}{{dx}} + x = c represents
View answer and explanation
Correct option: D
A family of circles whose centres are on the xx-axis
Option D (A family of circles whose centres are on the xx-axis) is correct because the given differential equation can be integrated to obtain the equation of a circle with the center on the x-axis.
273 Medium
273 of 452
Analyzing +4 | -1
The differential equation of the system of all circles of radius rr in the xyxy-plane, is
View answer and explanation
Correct option: C
[1+(dydx)2]3=r2(d2ydx2)2{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {r^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}
Option D ({\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {r^2}{\left( {\frac{{{{d^2}y}}{{d{x^2}}}} \right)^2}{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {r^2}{\left( {\frac{{{{d^2}y}}{{d{x^2}}}} \right)^2}) is correct because it represents the differential equation derived from the general equation of a circle with radius r.
274 Medium
274 of 452
Application +4 | -1
The solution of the differential equation xdydx=2y+x3exx\frac{{dy}}{{dx}} = 2y + {x^3}{e^x}, where y=0y = 0 when x=1x = 1, is
View answer and explanation
Correct option: C
y=x2(ex−e)y = {x^2}\left( {{e^x} - e} \right)
Option C (y=x2(ex−e)y = {x^2}\left( {{e^x} - e} \right)) is correct because it is the solution obtained by solving the linear differential equation and using the given initial condition.
275 Medium
275 of 452
Application +4 | -1
The general solution of the differential equation dydx+1+cos⁡2y1−cos⁡2x=0\frac{{dy}}{{dx}} + \frac{{1 + \cos 2y}}{{1 - \cos 2x}} = 0 is given by
View answer and explanation
Correct option: B
tan⁡y−cot⁡x=c\tan y - \cot x = c
Option B (tan⁡y−cot⁡x=c\tan y - \cot x = c) is correct because it is the general solution obtained by separating variables and integrating the given differential equation.
276 Medium
276 of 452
Application +4 | -1
Family of curves y=ex(Acos⁡x+Bsin⁡x)y=e^x(A \cos x+B \sin x), represents the differential equation
View answer and explanation
Correct option: B
d2ydx2=2dydx−2y\frac{d^2 y}{d x^2}=2 \frac{d y}{d x}-2 y
Option B (d2ydx2=2dydx−2y\frac{{d^2 y}}{{d x^2}}=2 \frac{{d y}}{{d x}}-2 y) is correct because differentiating the given family of curves twice and eliminating A and B leads to this differential equation.
277 Medium
277 of 452
Application +4 | -1
If x3dy+xydx=x2dy+2ydx;y(2)=ex^3 d y+x y d x=x^2 d y+2 y d x ; y(2)=e and x>1x>1, then y(4)y(4) is equal to
View answer and explanation
Correct option: B
32e\frac{3}{2} \sqrt{e}
Option B (−(d2ydx2)(dydx)−3-( \frac{d^2 y}{d x^2})( \frac{d y}{d x})^{-3}) is correct because it is the correct expression for the second derivative of x with respect to y, derived using the chain rule.
278 Medium
278 of 452
Application +4 | -1
Integral curve satisfying y′=x2+y2x2−y2,y(1)=2y' = \frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}},y\left( 1 \right) = 2 has the slope at the point (1,0) of the curve is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because the slope of the integral curve at the point (1,0) is found by substituting x=1 and y=0 into the derivative equation.
279 Medium
279 of 452
Application +4 | -1
The solution of (x+y+1)dydx=1\left( {x + y + 1} \right)\frac{{dy}}{{dx}} = 1 is
View answer and explanation
Correct option: C
x=−(y+2)+ceyx = - \left( {y + 2} \right) + c{e^y}
Option C, x=−(y+2)+ceyx = - \left( {y + 2} \right) + c{e^y}, is the correct solution because it satisfies the given differential equation after solving it as a linear differential equation in terms of x.
280 Medium
280 of 452
Application +4 | -1
If dx+dy=(x+y)(dx−dy)dx + dy = \left( {x + y} \right)\left( {dx - dy} \right), then log⁡(x+y)\log \left( {x + y} \right) is equal to
View answer and explanation
Correct option: C
x−y+cx - y + c
Option C (x−y+cx - y + c) is the correct solution because it is obtained by solving the given differential equation using substitution and integration.
281 Medium
281 of 452
Application +4 | -1
The equation of the curve satisfying the equation (xy−x2)dydx=y2\left( {xy - {x^2}} \right)\frac{{dy}}{{dx}} = {y^2} and passing through the point (−1,1)\left( { - 1,1} \right) is
View answer and explanation
Correct option: A
y=(log⁡y−1)xy = \left( {\log y - 1} \right)x
Option A (y=(log⁡y−1)xy = \left( {\log y - 1} \right)x) is the correct solution because it satisfies the given differential equation and passes through the point (-1, 1).
282 Medium
282 of 452
Application +4 | -1
The integrating factor of the differential equation xlog⁡xdydx+y=2log⁡xx\log x\frac{{dy}}{{dx}} + y = 2\log x is given by
View answer and explanation
Correct option: B
log⁡x\log x
Option B (log⁡x\log x) is correct because it is the integrating factor obtained by rewriting the differential equation in standard form and then applying the integrating factor formula.
283 Medium
283 of 452
Application +4 | -1
For solution of dydx+ytan⁡x=sec⁡x,y(0)=1\frac{d y}{d x}+y \tan x=\sec x, y(0)=1, then y(π6)y\left(\frac{\pi}{6}\right) is equal to:
View answer and explanation
Correct option: B
1+32\frac{1+\sqrt{3}}{2}
Option B (1+32\frac{{1+\sqrt{{3}}}}{{2}}) is correct because it is the value of y(π6)y(\frac{{\pi}}{{6}}) obtained by solving the given differential equation with the initial condition.
284 Medium
284 of 452
Application +4 | -1
The solution of the differential equation e−x(y+1)dy+(cos⁡2x−sin⁡2x)ydx=0{e^{ - x}}\left( {y + 1} \right)dy + \left( {{{\cos }^2}x - \sin 2x} \right)y\mathrm{ }dx = 0 subjected to the condition that y=1y = 1 when x=0x = 0 is
View answer and explanation
Correct option: A
y+log⁡y+excos⁡2x=2y + \log y + {e^x}{\cos ^2}x = 2
Option A (y+log⁡y+excos⁡2x=2y + \log y + {e^x}{\cos ^2}x = 2) is correct because it is the solution of the given differential equation that satisfies the given initial condition.
285 Medium
285 of 452
Application +4 | -1
Solution of the differential equation {1x−y2(x−y)2}dx+{x2(x−y)2−1y}dy=0\left\{\frac{1}{x}-\frac{y^2}{(x-y)^2}\right\} d x+\left\{\frac{x^2}{(x-y)^2}-\frac{1}{y}\right\} d y=0 is
View answer and explanation
Correct option: A
ln⁡∣xy∣+xyx−y=c\ln \left|\frac{x}{y}\right|+\frac{x y}{x-y}=c
Option A (ln⁡∣xy∣+xyx−y=c\ln |\frac{{x}}{{y}}| + \frac{{xy}}{{x-y}} = c) is the correct solution obtained by separating variables and integrating the given differential equation.

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286 Medium
286 of 452
Application +4 | -1
The differential equation of the family of circles with fixed radius 5 unit and centre on the line y=2y = 2, is
View answer and explanation
Correct option: D
(y−2)2y′2=25−(y−2)2{\left( {y - 2} \right)^2}{y'^2} = 25 - {\left( {y - 2} \right)^2}
Option D ((y−2)2y′2=25−(y−2)2{\left( {y - 2} \right)^2}{y'^2} = 25 - {\left( {y - 2} \right)^2}) is correct because it represents the differential equation derived from the family of circles with radius 5 and center on the line y=2.
287 Medium
287 of 452
Application +4 | -1
The solution of the differential equation (1+y2)tan⁡−1xdx+y(1+x2)dy=0\left( {1 + {y^2}} \right){\tan ^{ - 1}}xdx + y\left( {1 + {x^2}} \right)dy = 0 is
View answer and explanation
Correct option: B
log⁡(1+y2)+(tan⁡−1x)2=c\log \left( {1 + {y^2}} \right) + {\left( {{{\tan }^{ - 1}}x} \right)^2} = c
Option B (log⁡(1+y2)+(tan⁡−1x)2=c\log (1 + {y^2}) + {({{\tan }^{ - 1}}x)^2} = c) is correct because it is the solution obtained after separating variables and integrating the differential equation.
288 Medium
288 of 452
Application +4 | -1
Let y=y(x),y>0y=y(x), y>0, be a solution curve of the differential equation (1+x2)dy=y(x−y)dx\left(1+x^{2}\right) d y=y(x-y) d x. If y(0)=1y(0)=1 and y(22)=βy(2 \sqrt{2})=\beta, then
View answer and explanation
Correct option: A
e3β−1=e(3+22)e^{3 \beta^{-1}}=e(3+2 \sqrt{2})
Option A (e3β−1=e(3+22)e^{3 \beta^{-1}}=e(3+2 \sqrt{{2}})) is correct because it represents the solution to the given differential equation with the specified initial condition and value at x=22x = 2\sqrt{2}.
289 Medium
289 of 452
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.
290 Medium
290 of 452
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 ...... .
View answer and explanation
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.
291 Medium
291 of 452
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.
292 Medium
292 of 452
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.
293 Medium
293 of 452
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 .
View answer and explanation
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).
294 Medium
294 of 452
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.
295 Medium
295 of 452
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.
296 Medium
296 of 452
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.
297 Medium
297 of 452
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.
298 Medium
298 of 452
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.
299 Medium
299 of 452
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.
300 Medium
300 of 452
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.
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