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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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331 Medium
331 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 correct because it is the solution obtained after solving the given homogeneous differential equation by substitution and integration.
332 Easy
332 of 975
Application +4 | -1
The interval contained in the domain of definition of non-zero solutions of the differential equation (x−3)2y′+y=0(x-3)^2 y^{\prime}+y=0, is
View answer and explanation
Correct option: A
(−π/2,π/2)(-\pi / 2, \pi / 2)
Option A (−1-1) is correct because it is the value of y(log⁡e13)y(\log_e 13) obtained by solving the differential equation with the given initial condition.
333 Medium
333 of 975
Application +4 | -1
The curve satisfying the differential equation dydx=y(x+y3)x(y3−x)\frac{d y}{d x}=\frac{y\left(x+y^3\right)}{x\left(y^3-x\right)} and passing through (4,−2)(4,-2) is
View answer and explanation
Correct option: B
y=−2xy=-2 x
Option C, y3=−2xy^3=-2 x, is the correct curve obtained by solving the given differential equation and applying the given condition (4,-2).
334 Hard
334 of 975
Application +4 | -1
A rumour spreeds through a population of 5000 people at a rate proportional to the product of the number of people who have heard it and the number who have not. Suppose that 100 people initiate the rumour and that a total of 500 people know the rumour after 2 days. How long will it take for half the people to hear the rumour. (log⁡9/log⁡49=129/229)(\log 9/\log 49=129 / 229).
View answer and explanation
Correct option: A
5
Option A (5) is correct because after solving the differential equation and using the given conditions, it takes approximately 5 days for half the people to hear the rumor.
335 Easy
335 of 975
Understanding +4 | -1
The degree and the order of y=xd2ydx2+31+(d3ydx3)3\left.y=x \frac{d^2 y}{d x^2}+3 \sqrt{1+\left(\frac{d^3 y}{d x^3}\right.}\right)^3 are and respectively.
View answer and explanation
Correct option: B
3,3
Option B (3,3) is correct because after removing the radical, the highest order derivative is 3 and its power is 3.
336 Easy
336 of 975
Application +4 | -1
A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half its moisture during the first hour, then the time when it would have lost 99.9%99.9 \% of its moisture is (weather conditions remaining same)
View answer and explanation
Correct option: C
approximately 10 h10 \mathrm{~h}
Option C (approximately 10 h) is correct because it is derived from solving the differential equation and applying the given conditions to find the time when 99.9% of the moisture is lost.
337 Easy
337 of 975
Understanding +4 | -1
The solution of dydx+p(x)y=0\frac{d y}{d x}+p(x) y=0, is
View answer and explanation
Correct option: B
y=Ce−∫Pdxy=C e^{-\int P d x}
Option B (y=Ce−∫Pdxy=C e^{-\int P d x}) is the correct solution to the given differential equation after separation of variables and integration.
338 Medium
338 of 975
Application +4 | -1
General solution of the differential equation (cos⁡2x)dydx+y=tan⁡x   (0≤x<π2)\left(\cos ^2 x\right) \frac{d y}{d x}+y=\tan x \,\,\,\left(0 \leq x < \frac{\pi}{2}\right) is
View answer and explanation
Correct option: A
y+1=tan⁡x+ce−tan⁡xy+1=\tan x+c e^{-\tan x}
Option A (y+1=tan⁡x+ce−tan⁡xy+1=\tan x+c e^{-\tan x}) is correct because it is the solution obtained after solving the given linear differential equation.
339 Medium
339 of 975
Application +4 | -1
The solution of y5x+y−xdy dx=0y^5 x+y-x \frac{\mathrm{d} y}{\mathrm{~d} x}=0 is
View answer and explanation
Correct option: B
x5/5+(1/4)(x/y)4=Cx^5 / 5+(1 / 4)(x / y)^4=C
Option B (x5/5+(1/4)(x/y)4=Cx^5 / 5+(1 / 4)(x / y)^4=C) is correct because it is the solution obtained by multiplying the given differential equation by an appropriate integrating factor and then integrating.
340 Easy
340 of 975
Application +4 | -1
The solution of the equation dydx=sec⁡(x+y)\frac{d y}{d x}=\sec (x+y) is given by
View answer and explanation
Correct option: C
y−tan⁡12(x+y)=Cy-\tan \frac{1}{2}(x+y)=C
The correct answer is C (y−tan⁡12(x+y)=Cy-\tan \frac{1}{2}(x+y)=C) because it is the solution obtained by substituting x+y=ux+y=u and integrating the resulting equation.
341 Easy
341 of 975
Understanding +4 | -1
The degree of differential equation [1+(dydx)2]3/2=d2ydx2\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}=\frac{d^2 y}{d x^2} is
View answer and explanation
Correct option: D
2
Option D (2) is correct because after squaring both sides to remove the fractional exponent, the power of the highest order derivative becomes 2.
342 Medium
342 of 975
Understanding +4 | -1
The order and degree of the differential equation (1+3dydx)23=4d3ydx3{\left( {1 + 3\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^3}y}}{{d{x^3}}} are
View answer and explanation
Correct option: C
(3,3)\left( {3,3} \right)
Option C (3,33,3) is correct because after raising both sides to the power of 3/2, the highest order derivative is d3ydx3\frac{{{d^3}y}}{{d{x^3}}} (order 3) and its power is 3 (degree 3).
343 Medium
343 of 975
Analyzing +4 | -1
The differential equation representing the family of ellipse having foci either on the xx-axis or on the yy-axis centre at the origin and passing through the point (0,3)(0,3) is
View answer and explanation
Correct option: C
xyy′′+x(y′)2−yy′=0x y y^{\prime \prime}+x\left(y^{\prime}\right)^2-y y^{\prime}=0
Option C (xyy′′+x(y′)2−yy′=0x y y^{\prime\prime}+x(y^{\prime})^2-y y^{\prime}=0) is correct because it represents the differential equation of the family of ellipses centered at the origin, passing through (0,3), and having foci on either the x-axis or y-axis.
344 Easy
344 of 975
Understanding +4 | -1
Which of the following equation is linear
View answer and explanation
Correct option: B
x2dydx+y=exx^2 \frac{d y}{d x}+y=e^x
Option B (x2dydx+y=exx^2 \frac{{d y}}{{d x}}+y=e^x) is linear because it can be written in the standard form of a first-order linear differential equation: dydx+P(x)y=Q(x)\frac{{dy}}{{dx}} + P(x)y = Q(x).
345 Easy
345 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (x2+4)2dy+(2x3y+8xy−2)dx=0\left(x^2+4\right)^2 d y+\left(2 x^3 y+8 x y-2\right) d x=0. If y(0)=0y(0)=0, then y(2)y(2) is equal to
View answer and explanation
Correct option: D
π32\frac{\pi}{32}
Option D (π32\frac{\pi}{32}) is correct because it is the solution to the given differential equation with the specified initial condition, evaluated at x=2.
346 Easy
346 of 975
Application +4 | -1
The sum of the squares of the perpendicular drawn from the points (0,1)(0,1) and (0,−1)(0,-1) to any tangent to a curve is 2. The equation of the curve, is
View answer and explanation
Correct option: B
y=C(x±1)y=C(x \pm 1)
Option B, y=C(x±1)y=C(x \pm 1), is correct because it represents the solution to the differential equation derived from the given condition about the sum of squares of perpendiculars.
347 Easy
347 of 975
Application +4 | -1
If y(x)y(x) satisfies the differential equation y′−ytan⁡x=2xsec⁡xy^{\prime}-y \tan x=2 x \sec x and y(0)=0y(0)=0, then
View answer and explanation
Correct option: A
y(π4)=π282,y′(π3)=4π3+2π233y\left(\frac{\pi}{4}\right)=\frac{\pi^2}{8 \sqrt{2}}, y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^2}{3 \sqrt{3}}
Option A (y(π4)=π282,y′(π3)=4π3+2π233y(\frac{{\pi}}{{4}})=\frac{{\pi^2}}{{8 \sqrt{{2}}}}, y^{\prime}(\frac{{\pi}}{{3}})=\frac{{4 \pi}}{{3}}+\frac{{2 \pi^2}}{{3 \sqrt{{3}}}}) is correct because it provides the correct values of y and y' at the specified points, obtained by solving the given differential equation with the initial condition y(0) = 0.
348 Medium
348 of 975
Application +4 | -1
A curve having the condition that the slope of tangent at some point is two times the slope of the straight line joining the same point to the origin of coordinates, is a/an
View answer and explanation
Correct option: C
Parabola
The correct answer is C (Parabola) because the given condition leads to a differential equation whose solution is of the form y = cx², which represents a parabola.
349 Medium
349 of 975
Application +4 | -1
General solution of the equation (1+x2)dydx+xy=x3y3\left(1+x^2\right) \frac{d y}{d x}+x y=x^3 y^3 is
View answer and explanation
Correct option: B
1x2+1(1y2+1)=−1log⁡(1+x2)+c\frac{1}{x^2+1}\left(\frac{1}{y^2}+1\right)=-1 \log \left(1+x^2\right)+c
Option B (1x2+1(1y2+1)=−1log⁡(1+x2)+c\frac{1}{x^2+1}(\frac{1}{y^2}+1)=-1 \log (1+x^2)+c) is the correct general solution obtained by solving the Bernoulli differential equation.
350 Medium
350 of 975
Application +4 | -1
The solution of dυdt+kmυ=−g\frac{{d\upsilon }}{{dt}} + \frac{k}{m}\upsilon = - g is
View answer and explanation
Correct option: A
υ=ce−kmt−mgk\upsilon = c{e^{ - \frac{k}{m}t}} - \frac{{mg}}{k}
Option A (υ=ce−kmt−mgk\upsilon = c{e^{ - \frac{k}{m}t}} - \frac{{mg}}{k}) is correct because it is the solution to the given first-order linear differential equation.
351 Medium
351 of 975
Analyzing +4 | -1
If y=(x+1+x)ny = {\left( {x + \sqrt {1 + x} } \right)^n}, then (1+x2)d2ydx2+xdydx\left( {1 + {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} is
View answer and explanation
Correct option: A
n2y{n^2}y
Option A (n2y{n^2}y) is correct because after differentiating the given equation twice and simplifying, we obtain the expression equal to n2y{n^2}y.
352 Medium
352 of 975
Analyzing +4 | -1
If x=sec⁡θ−cos⁡θx=\sec \theta-\cos \theta and y=sec⁡nθ−cos⁡nθy=\sec ^n \theta-\cos ^n \theta, then (x2+4)(dydx)2\left(x^2+4\right)\left(\frac{d y}{d x}\right)^2 is equal to
View answer and explanation
Correct option: C
n2(y2+4)n^2\left(y^2+4\right)
The correct answer is C (n2(y2+4)n^2\left(y^2+4\right)) because it is the simplified expression obtained after differentiating xx and yy with respect to θ\theta and substituting into the given equation.
353 Easy
353 of 975
Application +4 | -1
Let the solution curve y=f(x)y=f(x) of the differential equation dydx+xyx2−1=x4+2x1−x2,x∈(−1,1)\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}}, x \in(-1,1) \quad pass through the origin. Then ∫−3232f(x)dx\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x is
View answer and explanation
Correct option: B
π3−34\frac{\pi}{3}-\frac{\sqrt{3}}{4}
Option B (π3−34\frac{\pi}{3}-\frac{\sqrt{3}}{4}) is correct because it is the value of the definite integral of the solution to the given differential equation, evaluated between the specified limits.
354 Medium
354 of 975
Application +4 | -1
The solution of the differential equation dydx=x−2y+12x−4y\frac{{dy}}{{dx}} = \frac{{x - 2y + 1}}{{2x - 4y}} is
View answer and explanation
Correct option: A
(x−2y)2+2x=c{\left( {x - 2y} \right)^2} + 2x = c
Option A ((x−2y)2+2x=c{\left( {x - 2y} \right)^2} + 2x = c) is correct because it is the solution to the differential equation obtained by using a suitable substitution.
355 Medium
355 of 975
Application +4 | -1
The solution of differential equation dtdx=t(ddx(g(x)))−t2g(x)\frac{{dt}}{{dx}} = \frac{{t\left( {\frac{d}{{dx}}\left( {\mathrm{g}\left( x \right)} \right)} \right) - {t^2}}}{{\mathrm{g}\left( x \right)}} is
View answer and explanation
Correct option: C
t=g(x)x+ct = \frac{{\mathrm{g}\left( x \right)}}{{x + c}}
Option D (t=g(x)x+ct = \frac{{\mathrm{g}\left( x \right)}}{{x + c}}) is correct because it is the solution to the given differential equation after applying a suitable substitution and integrating factor.
356 Easy
356 of 975
Application +4 | -1
The solution of y′′−4y′+3y=0,y(0)=6y^{\prime \prime}-4 y^{\prime}+3 y=0, y(0)=6 and y′(0)=10y^{\prime}(0)=10 is
View answer and explanation
Correct option: A
y=4ex+2e3xy=4 e^x+2 e^{3 x}
Option A (y=4ex+2e3xy=4 e^x+2 e^{3 x}) is correct because it satisfies the given differential equation and the initial conditions y(0)=6 and y'(0)=10.
357 Easy
357 of 975
Understanding +4 | -1
The differential equation of all ‘Simple Harmonic Motions’ of given period 2πn\frac{{2\pi }}{n}, is
View answer and explanation
Correct option: B
d2xdt2+n2x=0\frac{{{d^2}x}}{{d{t^2}}} + {n^2}x = 0
Option B (d2xdt2+n2x=0\frac{{{d^2}x}}{{d{t^2}}} + {n^2}x = 0) is correct because the general solution to this differential equation represents simple harmonic motion with period 2πn\frac{{2\pi }}{n}.
358 Easy
358 of 975
Application +4 | -1
The curve passing through (0,1)(0,1) and satisfying sin⁡(dydx)=12\sin \left(\frac{d y}{d x}\right)=\frac{1}{2} is
View answer and explanation
Correct option: B
sin⁡(y−1x)=12\sin \left(\frac{y-1}{x}\right)=\frac{1}{2}
Option B (sin⁡(y−1x)=12\sin(\frac{{y-1}}{{x}})=\frac{{1}}{{2}}) is correct because it satisfies the given differential equation and the initial condition.
359 Easy
359 of 975
Analyzing +4 | -1
Let dydx=ax−by+abx+cy+a,\frac{{dy}}{{dx}} = \frac{{ax - by + a}}{{bx + cy + a}}, where a,b,ca,b,c are constants, represents a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is
View answer and explanation
Correct option: B
8
Option B (8) is correct because after integrating the differential equation and using the given point to find the circle's equation, the shortest distance from the point (11, 6) to the circle is calculated to be 8.
360 Medium
360 of 975
Application +4 | -1
The solution of differential equation (xy2−ex13)dx−x2ydy=0\left(x y^2-e^{x^{\frac{1}{3}}}\right) d x-x^2 y d y=0 is
View answer and explanation
Correct option: B
3y2=2x2ex13+Cx23 y^2=2 x^2 e^{x^{\frac{1}{3}}}+C x^2
Option B (3y2=2x2ex13+Cx23 y^2=2 x^2 e^{x^{\frac{1}{3}}}+C x^2) is correct as it represents the solution to the given differential equation after using an integrating factor.
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