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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Medium

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Showing 181-210 of 452 questions

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181 Medium
181 of 452
Application +4 | -1
The differential equation dydx=1−y2y\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} determines a family 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 differential equation represents a family of circles with this property.
182 Medium
182 of 452
Application +4 | -1
The solution of dydx+1=cosec(x+y)\frac{{dy}}{{dx}} + 1 = \mathrm{cosec}\left( {x + y} \right) is
View answer and explanation
Correct option: A
cos⁡(x+y)+x=c\cos \left( {x + y} \right) + x = c
Option A (cos⁡(x+y)+x=c\cos \left( {x + y} \right) + x = c) is correct because it is the solution obtained by substituting x+y=tx+y=t and integrating the resulting equation.
183 Medium
183 of 452
Application +4 | -1
A curve passes through the point (1,π4)\left(1, \frac{\pi}{4}\right) and its slope at any point is given by yx−cos⁡2(yx)\frac{y}{x}-\cos ^2\left(\frac{y}{x}\right). Then, the curve has the equation
View answer and explanation
Correct option: A
y=xtan⁡−1(ln⁡ex)y=x \tan ^{-1}\left(\ln \frac{e}{x}\right)
Option A, y=xtan⁡−1(ln⁡ex)y=x \tan ^{-1}(\ln \frac{e}{x}), is the correct equation of the curve that satisfies the given differential equation and passes through the point (1,π4)(1, \frac{\pi}{4}).
184 Medium
184 of 452
Application +4 | -1
If the algebraic sub-tangent at any point of a curve is equal to half of the sum of the coordinates of the point, then the equation of the curve is
View answer and explanation
Correct option: B
(x−y)2=cy(x-y)^2=c y
Option C ((x−y)2=cx(x-y)^2=c x) is the correct answer because it satisfies the given condition about the algebraic sub-tangent after solving the differential equation.
185 Medium
185 of 452
Analyzing +4 | -1
If f(x)f(x) is a differentiable function in the interval ((0,∞)((0, \infty) such that f(a)=1f(a)=1 and lim⁡t→xt2f(x)−x2f(t)t−x=1\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1, for each x>0x > 0, then f(32)\mathrm{f}\left(\frac{3}{2}\right) is equal to:
View answer and explanation
Correct option: D
3118\frac{31}{18}
Option D (3118\frac{{31}}{{18}}) is correct because it is the value of f(3/2) obtained after solving the given differential equation and using the initial condition f(a) = 1.
186 Medium
186 of 452
Analyzing +4 | -1
The differential equation of the family of curves y=e2x(acos⁡x+bsin⁡x)y = {e^{2x}}\left( {a\cos x + b\sin x} \right), where aa and bb are arbitrary constants, is given by
View answer and explanation
Correct option: A
y2−4y1+5y=0{y_2} - 4{y_1} + 5y = 0
Option A (y2−4y1+5y=0{y_2} - 4{y_1} + 5y = 0) is correct because it represents the differential equation obtained after differentiating the given equation twice and eliminating the arbitrary constants a and b.
187 Medium
187 of 452
Application +4 | -1
Solution of cos⁡xdydx+ysin⁡x=1\cos x \frac{d y}{d x}+y \sin x=1 is
View answer and explanation
Correct option: B
ysec⁡xtan⁡x=cy \sec x \tan x=c
The correct answer is C (ytan⁡x=sec⁡x+cy \tan x=\sec x+c) because it is the solution obtained after finding the integrating factor and solving the linear differential equation.
188 Medium
188 of 452
Application +4 | -1
The equation of family of a curve is y2=4a(x+a){y^2} = 4a\left( {x + a} \right) then differential equation of the family is
View answer and explanation
Correct option: C
y=2y′x+yy′2y = 2y'x + y{y'^2}
Option C (y=2y′x+yy′2y = 2y'x + y{y'^2}) is the correct differential equation obtained by differentiating the given equation and eliminating the arbitrary constant 'a'.
189 Medium
189 of 452
Application +4 | -1
Let x=x(t)x=x(t) and y=y(t)y=y(t) be solutions of the differential equations dxdt+ax=0\frac{d x}{d t}+a x=0 and dydt+by=0\frac{d y}{d t}+b y=0 respectively, a,b∈R.a, b \in \mathbb{R}. Given that x(0)=2;y(0)=1x(0)=2 ; y(0)=1 and 3y(1)=2x(1),3 y(1)=2 x(1), the value of t,t, for which x(t)=y(t),x(t)=y(t), is:
View answer and explanation
Correct option: C
log⁡432\log _{\frac{4}{3}} 2
Option C is correct because it involves solving two first-order differential equations and then using the given condition 3y(1) = 2x(1) to find the value of t for which x(t) = y(t).
190 Medium
190 of 452
Application +4 | -1
If y=axn+1+bx−ny=a x^{n+1}+b x^{-n}, then x2d2ydx2x^2 \frac{d^2 y}{d x^2} equals to
View answer and explanation
Correct option: B
n(n+1)yn(n+1) y
Option B (n(n+1)yn(n+1) y) is correct because after differentiating twice and multiplying by x2x^2, the result is n(n+1)n(n+1) times the original function yy.
191 Medium
191 of 452
Analyzing +4 | -1
The differential equation of the family of parabolas with focus at the origin and the xx-axis as 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 correct because it represents the differential equation of the family of parabolas with focus at the origin and the x-axis as the axis.
192 Medium
192 of 452
Application +4 | -1
The solution of the differential equation (x2+y2)dx=2xydy\left( {{x^2} + {y^2}} \right)dx = 2xy\mathrm{ }dy is
(here cc is an arbitrary constant)
View answer and explanation
Correct option: B
c(x2−y2)=xc\left( {{x^2} - {y^2}} \right) = x
Option B (c(x2−y2)=xc\left( {{x^2} - {y^2}} \right) = x) is correct because it is the solution to the given homogeneous differential equation.
193 Medium
193 of 452
Application +4 | -1
Solution of the differential equation (xsin⁡2(yx)−y)dx+xdy=0,y=π4\left(x \sin ^2\left(\frac{y}{x}\right)-y\right) d x+x d y=0, y=\frac{\pi}{4} when x=1x=1 is
View answer and explanation
Correct option: A
cot⁡(yx)=log⁡(ex⁡)\cot \left(\frac{y}{x}\right)=\log (\operatorname{ex})
Option A, cot⁡(yx)=log⁡(ex⁡)\cot \left(\frac{y}{x}\right)=\log (\operatorname{ex}), is the correct solution obtained by solving the homogeneous differential equation and applying the given initial condition.
194 Medium
194 of 452
Application +4 | -1
Solution of dydx=yx+tan⁡yx\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x} is
View answer and explanation
Correct option: A
sin⁡(yx)=kx\sin \left(\frac{y}{x}\right)=k x
Option A (sin⁡(yx)=kx\sin \left(\frac{y}{x}\right)=k x) is correct because it is the solution obtained by substituting y=vxy=vx and integrating the resulting separable differential equation.
195 Medium
195 of 452
Application +4 | -1
The solution of x3dy dx+4x2x^3 \frac{\mathrm{d} y}{\mathrm{~d} x}+4 x^2 tan y=y= exsec⁡ye^x \sec y satisfying y(1)=0y(1)=0 is
View answer and explanation
Correct option: B
sin⁡y=ex(x−1)x−4\sin y=e^x(x-1) x^{-4}
Option B (sin⁡y=ex(x−1)x−4\sin y=e^x(x-1) x^{-4}) is correct because it is the solution to the given differential equation satisfying the initial condition y(1)=0.

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196 Medium
196 of 452
Application +4 | -1
The solution of (x−y3)dx+3xy2dy=0\left(x-y^3\right) d x+3 x y^2 d y=0 is
View answer and explanation
Correct option: B
log⁡x+y3x=k\log x+\frac{y^3}{x}=k
Option B (log⁡x+y3x=k\log x+\frac{y^3}{x}=k) is the correct solution obtained by solving the given differential equation using substitution and integration.
197 Medium
197 of 452
Application +4 | -1
The solution of the equation dydx+ytan⁡x=xmcos⁡x\frac{d y}{d x}+y \tan x=x^m \cos x is
View answer and explanation
Correct option: A
(m+1)y=xm+1cos⁡x+c(m+1)cos⁡x(m+1) y=x^{m+1} \cos x+c(m+1) \cos x
Option A ((m+1)y=xm+1cos⁡x+c(m+1)cos⁡x(m+1) y=x^{m+1} \cos x+c(m+1) \cos x) is the correct solution obtained by solving the linear differential equation using an integrating factor.
198 Medium
198 of 452
Application +4 | -1
The differential equation of the family of the curves x2+y2−2ax=0{x^2} + {y^2} - 2ax = 0 is
View answer and explanation
Correct option: B
y2−x2=2xyy′{y^2} - {x^2} = 2xyy'
Option B, y2−x2=2xyy′{y^2} - {x^2} = 2xyy', is the correct differential equation because it is obtained by differentiating the given equation and eliminating the arbitrary constant 'a'.
199 Medium
199 of 452
Application +4 | -1
If xdy−ydx+xcos⁡(log⁡ex)dx=0,y(1)=1x d y-y d x+x \cos \left(\log _e^x\right) d x=0, y(1)=1, then y(e)=y(e)=
View answer and explanation
Correct option: B
e(1−sin⁡1)e(1-\sin 1)
Option B (e(1-sin 1)) is correct because solving the differential equation and applying the initial condition y(1)=1 leads to y(e) = e(1-sin 1).
200 Medium
200 of 452
Analyzing +4 | -1
The degree of differential equation satisfying the relation 1+x2+1+y2=\sqrt{1+x^2}+\sqrt{1+y^2}= λ(x1+y2−y1+x2)\lambda\left(x \sqrt{1+y^2}-y \sqrt{1+x^2}\right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because after simplification and differentiation, the highest order derivative present is of order 1, and its power is also 1.
201 Medium
201 of 452
Application +4 | -1
Solution of the equation dydx=ex−y(ex−ey)\frac{d y}{d x}=e^{x-y}\left(e^x-e^y\right) is
View answer and explanation
Correct option: B
ey=(ex−1)+c(eex)−1e^y=\left(e^x-1\right)+c\left(e^{e^x}\right)^{-1}
Option B (ey=(ex−1)+c(eex)−1e^y=(e^x-1)+c(e^{e^x})^{-1}) is the correct solution to the given differential equation, obtained by transforming it into a linear form and applying the integrating factor.
202 Medium
202 of 452
Application +4 | -1
Let the population of rats surviving at time tt be govemed by the differential equation dp(t)dt=0.5p(t)−450\frac{d p(t)}{d t}=0.5 p(t)-450. If p(0)=850p(0)=850, then at what time will they extinct ?
View answer and explanation
Correct option: A
2log⁡182 \log 18
Option A (2log⁡182 \log 18) is correct because it represents the time at which the rat population becomes zero, calculated by solving the given differential equation.
203 Medium
203 of 452
Application +4 | -1
The temperature T(t)T(t) of a body at time t=0t=0 is 160∘F160^{\circ} \mathrm{F} and it decreases continuously as per the differential equation dTdt=−K(T−80)\frac{d T}{d t}=-K(T-80), where KK is a positive constant. If T(15)=120∘FT(15)=120^{\circ} \mathrm{F}, then T(45)T(45) is equal to
View answer and explanation
Correct option: B
90∘F90^{\circ} \mathrm{F}
Option B (90∘F90^{\circ} \mathrm{F}) is correct because it is the temperature obtained by solving the given differential equation with the provided initial conditions and time values.
204 Medium
204 of 452
Application +4 | -1
The solution of the differential equation dydx=xy+yxy+x\frac{{dy}}{{dx}} = \frac{{xy + y}}{{xy + x}} is
View answer and explanation
Correct option: D
y−x=log⁡(cxy)y - x = \log \left( {\frac{{cx}}{y}} \right)
Option D (y−x=log⁡(cxy)y - x = \log \left( {\frac{{cx}}{y}} \right)) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
205 Medium
205 of 452
Application +4 | -1
The real value of nn for which the substitution y=uny = {u^n} will transform the differential equation 2x4ydydx+y4=4x62{x^4}y\frac{{dy}}{{dx}} + {y^4} = 4{x^6} into a homogenous equation is
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C (32\frac{3}{2}) is correct because it is the value of n that transforms the given differential equation into a homogeneous equation.
206 Medium
206 of 452
Application +4 | -1
The slope of the tangent at (x,y)\left( {x,y} \right) to a curve passing through a point (2,1)\left( {2,1} \right) is x2+y22xy\frac{{{x^2} + {y^2}}}{{2xy}}, then the equation of the curve is
View answer and explanation
Correct option: A
2(x2−y2)=3x2\left( {{x^2} - {y^2}} \right) = 3x
Option A (2(x2−y2)=3x2\left( {{x^2} - {y^2}} \right) = 3x) is correct because it is the equation of the curve obtained by solving the differential equation and applying the given point (2, 1).
207 Medium
207 of 452
Analyzing +4 | -1
If (cos⁡xcot⁡y−sin⁡xtan⁡y)dy+(cos⁡xln⁡cos⁡y−sin⁡xln⁡sin⁡y)dx;0<y<π2(\cos x \cot y-\sin x \tan y) d y+(\cos x \ln \cos y-\sin x \ln \sin y) d x ; 0 < y < \frac{\pi}{2}. If y(π2)=π4y\left(\frac{\pi}{2}\right)=\frac{\pi}{4}, then the value of y(0)y(0) is
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C is correct because solving the given differential equation and applying the initial condition yields a constant, and subsequently, y(0) = π/4.
208 Medium
208 of 452
Application +4 | -1
A curve passes through (1,2)(1,2). If the product of the slope of the tangent at any point (x,y)(x, y) of the curve and yy-coordinate of the point is equal to its xx-coordinate, then the curve is
View answer and explanation
Correct option: C
A rectangular hyperbola
Option C (A rectangular hyperbola) is correct because the equation derived from the given condition represents a rectangular hyperbola.
209 Medium
209 of 452
Application +4 | -1
The solution of diferential 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 because it is the solution obtained after solving the given differential equation using an integrating factor.
210 Medium
210 of 452
Application +4 | -1
General solution of the equation dydx+2ytan⁡x=y2\frac{d y}{d x}+2 y \tan x=y^2 is
View answer and explanation
Correct option: C
cos⁡2x+y(x2+14sin⁡2x)=cy\cos ^2 x+y\left(\frac{x}{2}+\frac{1}{4} \sin 2 x\right)=c y
Option C (cos⁡2x+y(x2+14sin⁡2x)=cy\cos ^2 x+y(\frac{x}{2}+\frac{1}{4} \sin 2 x)=c y) is the correct general solution obtained by solving the Bernoulli differential equation.
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