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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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Showing 631-660 of 975 questions

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631 Easy
631 of 975
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.
632 Easy
632 of 975
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.
633 Easy
633 of 975
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.
634 Easy
634 of 975
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.
635 Easy
635 of 975
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}}.
636 Medium
636 of 975
Application +4 | -1
If y=y(x)y=y(x) is the solution curve of the differential equation (x2−4)dy−(y2−3y)dx=0,\left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0, x>2,x>2, y(4)=32y(4)=\frac{3}{2} and the slope of the curve is never zero, then the value of y(10)y(10) equals:
View answer and explanation
Correct option: A
31+(8)14\frac{3}{1+(8)^{\frac{1}{4}}}
Option A (31+(8)14\frac{3}{1+(8)^{\frac{1}{4}}}) is correct because it is the value of y(10) obtained after solving the differential equation and applying the given initial condition.
637 Medium
637 of 975
Application +4 | -1
The solution of the differential equation ydx+xdyydx−xdy=x2exyy4\frac{y d x+x d y}{y d x-x d y}=\frac{x^2 e^{x y}}{y^4} satisfying y(0)=1y(0)=1, is
View answer and explanation
Correct option: B
x3=3y3(1−e−xy)x^3=3 y^3\left(1-e^{-x y}\right)
Option B (x3=3y3(1-e(-xy))) is correct because it is the solution to the differential equation that satisfies the initial condition y(0)=1.
638 Medium
638 of 975
Application +4 | -1
A particular solution of log⁡dy dx=4x+5y,y(0)=0\log \frac{\mathrm{d} y}{\mathrm{~d} x}=4 x+5 y, \quad y(0)=0 is given by
View answer and explanation
Correct option: A
5e4x+4e−5y=95 e^{4 x}+4 e^{-5 y}=9
The correct answer is A (5e4x+4e−5y=95 e^{4 x}+4 e^{-5 y}=9) because it is the particular solution to the given differential equation with the initial condition y(0)=0.
639 Easy
639 of 975
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.
640 Easy
640 of 975
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.
641 Medium
641 of 975
Application +4 | -1
The differential equation dydx=1−y2y\frac{{dy}}{{dx}} = \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 general solution represents a family of circles with radius 1 and centers on the x-axis.
642 Medium
642 of 975
Application +4 | -1
The integrating factor of the differential equation dydx+y(1−x)x=1−x\frac{{dy}}{{dx}} + \frac{y}{{\left( {1 - x} \right)\sqrt x }} = 1 - \sqrt x is
View answer and explanation
Correct option: B
1+x1−x\frac{{1 + \sqrt x }}{{1 - \sqrt x }}
Option B (1+x1−x\frac{{1 + \sqrt x }}{{1 - \sqrt x }}) is correct because it is the integrating factor of the given differential equation.
643 Hard
643 of 975
Application +4 | -1
Solution of differential equation dydx=y−xy+x\frac{d y}{d x}=\frac{y-x}{y+x} is
View answer and explanation
Correct option: A
log⁡e(x2+y2)+2tan⁡−1yx+c=0\log _e\left(x^2+y^2\right)+2 \tan ^{-1} \frac{y}{x}+c=0
Option A (log⁡e(x2+y2)+2tan⁡−1yx+c=0\log _e(x^2+y^2)+2 \tan ^{-1} \frac{y}{x}+c=0) is the correct solution obtained by solving the given homogeneous differential equation using the substitution y=vxy=vx.
644 Medium
644 of 975
Application +4 | -1
The general solution of the differential equation xdy+ydx+2x3dx=0x d y+y d x+2 x^3 d x=0, is
View answer and explanation
Correct option: B
xy+12x4=Cx y+\frac{1}{2} x^4=C
Option B (xy+12x4=Cx y+\frac{1}{2} x^4=C) is correct because it is the general solution obtained by integrating the given differential equation.
645 Medium
645 of 975
Application +4 | -1
For y=cos⁡kxy=\cos k x to be a solution of differential equation d2ydx2+4y=0\frac{d^2 y}{d x^2}+4 y=0, the value of kk is
View answer and explanation
Correct option: A
2
Option A (2) is correct because substituting y = cos(kx) into the differential equation and solving for k yields k = 2.

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646 Easy
646 of 975
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.
647 Easy
647 of 975
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.
648 Medium
648 of 975
Application +4 | -1
The solution of the differential equation x+ydydx=2yx+y \frac{d y}{d x}=2 y is
View answer and explanation
Correct option: B
log⁡(y−x)=c+xy−x\log (y-x)=c+\frac{x}{y-x}
Option B (log⁡(y−x)=c+xy−x\log (y-x)=c+\frac{x}{y-x}) is the correct solution of the given differential equation, which is solved by substituting y=vxy=vx.
649 Medium
649 of 975
Application +4 | -1
The equation of the curve in which the portion of yy-axis cut off between the origin and the tangent varies as the cube of the abscissa of the point of contact is
View answer and explanation
Correct option: C
y=−kx32+cxy = - \frac{{k{x^3}}}{2} + c\mathrm{ }x
Option C (y=−kx32+cxy = - \frac{{k{x^3}}}{2} + c\mathrm{ }x) is correct because it is the solution to the differential equation formed based on the given condition about the y-intercept of the tangent.
650 Easy
650 of 975
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.
651 Easy
651 of 975
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.
652 Easy
652 of 975
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.
653 Medium
653 of 975
Application +4 | -1
The general solution of the differential equation ydy+1+y2dx=0y d y+\sqrt{1+y^2} d x=0 represents a family of
View answer and explanation
Correct option: C
Hyperbolas
Option C (Hyperbolas) is correct because the general solution of the given differential equation represents a family of hyperbolas.
654 Easy
654 of 975
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.
655 Easy
655 of 975
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.
656 Hard
656 of 975
Application +4 | -1
If the curve y(x)y(x) satisfying xdx=(x2y−y3)dyx d x=\left(\frac{x^2}{y}-y^3\right) d y passes through (0,2)(0,2), then value of (y(4))2(4−(y(4))2)(y(4))^2\left(4-(y(4))^2\right) is
View answer and explanation
Correct option: A
16
Option A (16) is correct. The solution to the differential equation is found, and the given point is used to find the constant of integration. Finally, y(4) is calculated and substituted into the expression.
657 Medium
657 of 975
Application +4 | -1
The 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 is the solution obtained after separating variables and integrating the differential equation.
658 Easy
658 of 975
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.
659 Easy
659 of 975
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.
660 Easy
660 of 975
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).
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