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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Easy

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Showing 301-330 of 494 questions

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301 Easy
301 of 494
Application +4 | -1
If xpyq=(x+y)p+qx^p y^q=(x+y)^{p+q}, then dy/dxd y / d x is equal to
View answer and explanation
Correct option: A
yx\frac{y}{x}
Option A (yx\frac{y}{x}) is correct because by taking the logarithm of both sides and differentiating implicitly, we find that dydx=yx\frac{dy}{dx} = \frac{y}{x}.
302 Easy
302 of 494
Application +4 | -1
The solution curve of the differential equation, (1+e−x)(1+y2)dydx=y2(1 + {e^{ - x}})(1 + {y^2})\frac{{dy}}{{dx}} = {y^2}, which passes through the point (0, 1) is
View answer and explanation
Correct option: C
y2=1+ylog⁡e[1+ex2]{y^2} = 1 + y{\log _e}\left[ {\frac{{1 + {e^x}}}{2}} \right]
Option C (y2=1+ylog⁡e[1+ex2]{y^2} = 1 + y{\log _e}\left[ {\frac{{1 + {e^x}}}{2}} \right]) is correct because it represents the solution to the given differential equation that passes through the point (0, 1).
303 Easy
303 of 494
Application +4 | -1
The general solution of the differential equation (1+y2)dx+(1+x2)dy=0\left( {1 + {y^2}} \right)dx + \left( {1 + {x^2}} \right)dy = 0 is
View answer and explanation
Correct option: C
x+y=c(1−xy)x + y = c\left( {1 - xy} \right)
Option D (x+y=c(1−xy)x + y = c(1 - xy)) is correct because it is the general solution obtained by separating variables and integrating the given differential equation.
304 Easy
304 of 494
Application +4 | -1
Let us consider a curve, y=f(x)y = f(x) passing through the point (−2,  2)( - 2,\,\,2) and the slope of the tangent to the curve at any point (x,f(x))(x,f(x)) is given by f(x)+xf′(x) =x2f(x) + xf'(x)\, = {x^2} then
View answer and explanation
Correct option: C
x3−3xf(x)−4=0{x^3} - 3x\mathrm{ }f(x) - 4 = 0
The correct answer is x3−3xf(x)−4=0{x^3} - 3x\mathrm{ }f(x) - 4 = 0 because it is obtained by solving the given differential equation and using the given point to find the constant of integration.
305 Easy
305 of 494
Application +4 | -1
A differential equation associated with the primitive y=a+be5x+ce−7xy=a+b e^{5 x}+c e^{-7 x} is
View answer and explanation
Correct option: C
y3+2y2−35y1=0y_3+2 y_2-35 y_1=0
Option C (y3+2y2−35y1=0y_3+2 y_2-35 y_1=0) is correct because it is the differential equation associated with the given primitive.
306 Easy
306 of 494
Application +4 | -1
If y(t)y(t) is a solution of (1+t)dydt−ty=1(1+t) \frac{d y}{d t}-t y=1 and y(0)=−1y(0)=-1, then y(1)y(1) is equal to
View answer and explanation
Correct option: A
−12-\frac{1}{2}
Option A (-12\frac{{1}}{{2}}) is correct because, after solving the differential equation and applying the initial condition, y(1) equals -12\frac{{1}}{{2}}.
307 Easy
307 of 494
Application +4 | -1
If the integrating factor of the differential equation dydx+P(x)y=Q(x)\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) is xx, then P(x)P\left( x \right) is
View answer and explanation
Correct option: C
1x\frac{1}{x}
Option C (1x\frac{1}{x}) is correct because the integrating factor is given as x, and P(x) can be found by differentiating the logarithm of the integrating factor.
308 Easy
308 of 494
Application +4 | -1
The solution of the differential equation dydx+1=ex+y\frac{d y}{d x}+1=e^{x+y}, is
View answer and explanation
Correct option: D
(x−C)ex+y+1=0(x-C) e^{x+y}+1=0
Option D ((x−C)ex+y+1=0(x-C) e^{x+y}+1=0) is correct because it is the solution obtained by solving the given differential equation using a suitable substitution.
309 Easy
309 of 494
Application +4 | -1
If the solution of differential equation x2d2ydx2+2xdydx=12yx^2 \frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}=12 y is y=Axm+Bx−ny=A x^m+B x^{-n} then find the value of m+nm+n, if m & n∈Nm \,\&\, n \in N.
View answer and explanation
Correct option: A
7
Option A (7) is correct because by substituting the given solution into the differential equation and solving for m and n, we find that m+n = 7.
310 Easy
310 of 494
Application +4 | -1
The solution of the equation (x+y)dy−(x−y)(x+y) \mathrm{d} y-(x-y) dx=0\mathrm{d} x=0 is
View answer and explanation
Correct option: C
x3+2xy−y2=Kx^3+2 x y-y^2=K
The correct answer is not listed in the options. The closest option is B (y2+2xy−x2=Ky^2+2 x y-x^2=K), but it is incorrect. The correct solution is x2−2xy−y2=Kx^2 - 2xy - y^2 = K.
311 Easy
311 of 494
Application +4 | -1
A particular solution of log⁡dy dx=3x+\log \frac{\mathrm{d} y}{\mathrm{~d} x}=3 x+ 4y,y(0)=04 y, y(0)=0 is
View answer and explanation
Correct option: D
4e3x+3e−4y=74 e^{3 x}+3 e^{-4 y}=7
Option D (4e3x+3e−4y=74 e^{3 x}+3 e^{-4 y}=7) is correct because it is the particular solution obtained by solving the separable differential equation and applying the initial condition y(0)=0.
312 Easy
312 of 494
Application +4 | -1
The solution of initial value problem y′=−2xyy^{\prime}=-2 x y, y(0)=1y(0)=1 represents
View answer and explanation
Correct option: C
bell-shaped curve
Option C (bell-shaped curve) is correct because the solution to the initial value problem is a Gaussian function, which is bell-shaped.
313 Easy
313 of 494
Application +4 | -1
Solution of differential equation (2xcos⁡y+y2cos⁡x)dx+(2ysin⁡x−x2sin⁡y)dy=0\left(2 x \cos y+y^{2} \cos x\right) d x+\left(2 y \sin x-x^{2} \sin y\right) d y=0 is
View answer and explanation
Correct option: A
x2cos⁡y+y2sin⁡x=cx^{2} \cos y+y^{2} \sin x=c
Option A is correct because the solution is found by recognizing the exact differential equation and integrating.
314 Easy
314 of 494
Application +4 | -1
The solution of the differential equation xdy−ydx−x2−y2dx=0x\mathrm{ }dy - y\mathrm{ }dx - \sqrt {{x^2} - {y^2}} dx = 0 is
View answer and explanation
Correct option: B
y+x2+y2=cx2y + \sqrt {{x^2} + {y^2}} = c{x^2}
Option B (y+x2+y2=cx2y + \sqrt {{x^2} + {y^2}} = c{x^2}) is correct because it is the solution obtained by solving the homogeneous differential equation using the substitution y = vx.
315 Easy
315 of 494
Understanding +4 | -1
Which of the following equation is non-linear
View answer and explanation
Correct option: B
ydydx+4x=0y \frac{d y}{d x}+4 x=0
Option B (ydydx+4x=0y \frac{{d y}}{{d x}}+4 x=0) is correct because it is non-linear due to the product of the dependent variable yy and its derivative dydx\frac{{d y}}{{d x}}.

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316 Easy
316 of 494
Application +4 | -1
The general solution excos⁡ydx−exsin⁡ydy=0{e^x}\cos ydx - {e^x}\sin ydy = 0, is
View answer and explanation
Correct option: D
excos⁡y=C{e^x}\cos y = C
Option D (excos⁡y=C{e^x}\cos y = C) is correct because the differential equation can be rearranged and integrated directly to obtain this solution.
317 Easy
317 of 494
Application +4 | -1
The general solution of the differential equation 1+x2+y2+x2y2+xydydx=0\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} + xy\frac{{dy}}{{dx}} = 0 is (where C is a constant of integration)
View answer and explanation
Correct option: A
1+y2+1+x2=12log⁡e(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = \frac{1}{2}{\log _e}\left( {\frac{{\sqrt {1 + {x^2}} + 1}}{{\sqrt {1 + {x^2}} - 1}}} \right) + C
The correct answer is A (1+y2+1+x2=12log⁡e(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = \frac{1}{2}{\log _e}\left( {\frac{{\sqrt {1 + {x^2}} + 1}}{{\sqrt {1 + {x^2}} - 1}}} \right) + C) because it represents the general solution obtained after separating variables and integrating the given differential equation.
318 Easy
318 of 494
Application +4 | -1
The solution of the differential equation dydx=x+yx\frac{d y}{d x}=\frac{x+y}{x} satisfying the condition y(1)=1y(1)=1 is
View answer and explanation
Correct option: D
y=xlog⁡x+xy=x \log x+x
Option D (y=xlog⁡x+xy=x \log x+x) is correct because it is the solution to the given differential equation that satisfies the initial condition y(1)=1y(1)=1.
319 Easy
319 of 494
Application +4 | -1
The differential equation of the curve y=e−x+ax+by=e^{-x}+a x+b is
( a,ba, b are arbitrary constants)
View answer and explanation
Correct option: A
exd2ydx2−1e^x \frac{d^2 y}{d x^2}-1
The correct answer is exd2ydx2=1e^x \frac{{d^2y}}{{dx^2}} = 1 because it represents the differential equation of the given curve after differentiating twice and eliminating the arbitrary constants.
320 Easy
320 of 494
Application +4 | -1
The solution of dydx=ax+hby+k\frac{d y}{d x}=\frac{a x+h}{b y+k} may represent a hyperbola when
View answer and explanation
Correct option: A
a>0,b>0a>0, b>0
Option A (a>0,b>0a>0, b>0) is correct because the given differential equation represents a hyperbola when ab>0ab > 0.
321 Easy
321 of 494
Application +4 | -1
The order and degree of the differential equation y=xdydx+a2(dydx)2+b2y = x\frac{{dy}}{{dx}} + \sqrt {{a^2}{{\left( {\frac{{dy}}{{dx}}} \right)}^2} + {b^2}} is
View answer and explanation
Correct option: B
1,21,2
Option B (1,2) is correct because after rewriting the equation and squaring both sides to remove the square root, the highest order derivative is of order 1, and its power is 2.
322 Easy
322 of 494
Application +4 | -1
The equation of the family of curves which intersect the hyperbola xy=2x y=2 orthogonally is
View answer and explanation
Correct option: A
y=x36+Cy=\frac{x^3}{6}+C
Option A (y=x36+Cy=\frac{{x^3}}{{6}}+C) is correct because it represents the family of curves that intersect the given hyperbola orthogonally, derived by finding the negative reciprocal of the hyperbola's derivative.
323 Easy
323 of 494
Application +4 | -1
y=xx+1y=\frac{x}{x+1} is a solution of the differential equation
View answer and explanation
Correct option: B
x2dydx=y2x^2 \frac{d y}{d x}=y^2
Option B (x2dydx=y2x^2 \frac{{d y}}{{d x}}=y^2) is correct because the derivative of y=xx+1y=\frac{{x}}{{x+1}} satisfies this differential equation.
324 Easy
324 of 494
Application +4 | -1
Integrating factor of dydx+yx=x3−3\frac{d y}{d x}+\frac{y}{x}=x^3-3 is
View answer and explanation
Correct option: A
xx
Option A (xx) is correct because it is the integrating factor of the given linear differential equation.
325 Easy
325 of 494
Application +4 | -1
If the solution curve of the differential equation (2x−10y3)dy+ydx=0(2x - 10{y^3})dy + ydx = 0, passes through the points (0, 1)(0,\,1) and (2, β)(2,\,\beta ) then β\beta is a root of the equation
View answer and explanation
Correct option: D
y5−y2−1=0{y^5} - {y^2} - 1 = 0
Option D (y5−y2−1=0{y^5} - {y^2} - 1 = 0) is correct because it represents the equation that β\beta satisfies after solving the given differential equation and using the given points.
326 Easy
326 of 494
Application +4 | -1
If y=y(x)y=y(x) is the solution of the differential equation, xdy dx+2y=x2x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=x^2 satisfying y(1)=1y(1)=1, then y(12)y\left(\frac{1}{2}\right) is equal to:
View answer and explanation
Correct option: C
4916\frac{49}{16}
Option C (49/16) is correct because solving the differential equation and applying the initial condition yields y(1/2) = 49/16.
327 Easy
327 of 494
Application +4 | -1
The equation of the curve passing through the point (5,4)(5,4) if the sum of reciprocal of the intercepts of the normal drawn at any point P(x,y){P}({x}, {y}) on it is
View answer and explanation
Correct option: B
(x−1)2+(y−1)2=25(x-1)^2+(y-1)^2=25
Option B ((x−1)2+(y−1)2=25(x-1)^2+(y-1)^2=25) is correct because it is the particular solution to the differential equation derived from the given condition and passing through the point (5,4).
328 Easy
328 of 494
Application +4 | -1
dydx+ytan⁡x=sec⁡x\frac{d y}{d x}+y \tan x=\sec x has the solution ........
View answer and explanation
Correct option: A
y=sin⁡x+ccos⁡xy=\sin x+c \cos x
Option A (y=sin⁡x+ccos⁡xy = \sin x + c \cos x) is correct because it is the solution obtained by solving the linear differential equation using an integrating factor.
329 Easy
329 of 494
Application +4 | -1
The general solution of a differential equation dydx=ex+y\frac{d y}{d x}=e^{x+y} is :
View answer and explanation
Correct option: A
ex+e−y=Ce^x+e^{-y}=C
Option A (ex+e−y=Ce^x+e^{-y}=C) is correct because separating variables and integrating leads to this solution.
330 Easy
330 of 494
Application +4 | -1
The Solution of 1−x2sin⁡−1xdy−ydx=0\sqrt{1-x^2} \sin ^{-1} x d y-y d x=0
View answer and explanation
Correct option: A
y=csin⁡−1xy=c \sin ^{-1} x
Option A (y=csin⁡−1xy=c \sin ^{-1} x) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
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