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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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151 Medium
151 of 975
Application +4 | -1
A solution of the differential equation dy dx=1xy[x2sin⁡y2+1]\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{x y\left[x^2 \sin y^2+1\right]} is
( CC is an arbitrary constant)
View answer and explanation
Correct option: A
x2(cos⁡y2−sin⁡y2−2Ce−y2)=2x^2\left(\cos y^2-\sin y^2-2 C e^{-y^2}\right)=2
Option D (none of these) is correct because the provided solution options do not match the actual solution of the given differential equation after applying the appropriate substitutions and integration.
152 Easy
152 of 975
Application +4 | -1
The elimination of the arbitrary constant mm from the equation y=emxy=e^{m x} gives the differential equation
View answer and explanation
Correct option: C
dydx=(yx)log⁡y\frac{d y}{d x}=\left(\frac{y}{x}\right) \log y
Option C (dydx=(yx)log⁡y\frac{{d y}}{{d x}}=\left(\frac{{y}}{{x}}\right) \log y) is correct because eliminating 'm' from y=emxy=e^{mx} leads to this differential equation.
153 Medium
153 of 975
Application +4 | -1
If, x=Acos⁡4t+Bsin⁡4tx = A\cos 4t + B\sin 4t then d2xdt2\frac{{{d^2}x}}{{d{t^2}}} is equal to
View answer and explanation
Correct option: A
−16x- 16x
Option A (−16x- 16x) is correct because the second derivative of the given function is equal to -16 times the original function.
154 Easy
154 of 975
Understanding +4 | -1
The degree of the differential equation y32/3+2+3y2+y1=0y_3^{2/3} + 2 + 3{y_2} + {y_1} = 0, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because after raising to the power of 3 to eliminate the fractional exponent, the power of the highest order derivative becomes 2.
155 Easy
155 of 975
Application +4 | -1
General solutions of y′=ey/x+y/xy^{\prime}=e^{y / x}+y / x is
View answer and explanation
Correct option: B
log⁡∣Cx∣=−e−y/x\log |C x|=-e^{-y / x}
Option B (log⁡∣Cx∣=−e−y/x\log |C x|=-e^{-y / x}) is correct because it is the general solution to the given differential equation.
156 Easy
156 of 975
Understanding +4 | -1
Which of the following statements on ordinary differential equations is/are true?
(i) The number of arbitrary constants is same as the degree of the differential equation.
(ii) A linear differential equation can contain products of the dependent variable and its derivatives.
(iii) A particular integral cannot contain arbitrary constants.
(iv) By putting v=yxv=\frac{y}{x} any homogeneous first order differential equation transforms to variable separable form.
View answer and explanation
Correct option: C
(iii) only
Option C ((iii) only) is correct because only statement (iii) is true; a particular integral cannot contain arbitrary constants.
157 Medium
157 of 975
Application +4 | -1
The solution of the differential equation dydx=ycos⁡x−y2sin⁡x\frac{d y}{d x}=\frac{y \cos x-y^2}{\sin x} is equal to (where c{c} is an arbitrary constant)
View answer and explanation
Correct option: C
sin⁡x=xy+cy\sin x=x y+c y
Option C (sin⁡x=xy+cy\sin x=x y+c y) is correct because it is the solution obtained after rearranging and integrating the given differential equation.
158 Easy
158 of 975
Application +4 | -1
The differential equation of the family of lines through the origin, is
View answer and explanation
Correct option: A
xdydx−y=0x \frac{d y}{d x}-y=0
Option A (xdydx−y=0x \frac{d y}{d x}-y=0) is correct because it represents the differential equation of the family of lines through the origin.
159 Medium
159 of 975
Application +4 | -1
If sin⁡(yx)=log⁡e∣x∣+α2\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2} is the solution of the differential equation xcos⁡(yx)dydx=ycos⁡(yx)+xx \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x and y(1)=π3y(1)=\frac{\pi}{3}, then α2\alpha^2 is equal to
View answer and explanation
Correct option: C
3
Option C (3) is correct because after solving the differential equation and applying the given condition, the value of α2\alpha^2 is found to be 3.
160 Medium
160 of 975
Application +4 | -1
(x2+y2)dy=xydx\left(x^2+y^2\right) d y=x y d x. If y(x0)=e,y(1)=1y\left(x_0\right)=e, y(1)=1, then value of x0=x_0=
View answer and explanation
Correct option: A
3e\sqrt{3} e
Option A (3e\sqrt{3} e) is the correct value of x0x_0, found by solving the differential equation and applying the given conditions.
161 Medium
161 of 975
Analyzing +4 | -1
The number of solutions of y′=y+1x−1,y(1)=2y' = \frac{{y + 1}}{{x - 1}},y\left( 1 \right) = 2 is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the given initial condition y(1) = 2 contradicts the general solution obtained from the differential equation, which requires y(1) = -1.
162 Medium
162 of 975
Application +4 | -1
If y=y(x)y=y(x) is a function of xx and y(1)=0y(1)=0, then the solution of the equation xdydx−yx+1=xx \frac{d y}{d x}-\frac{y}{x+1}=x is
View answer and explanation
Correct option: D
y=xx+1(x−1+log⁡∣x∣)y=\frac{x}{x+1}(x-1+\log |x|)
Option D (y=xx+1(x−1+log⁡∣x∣)y=\frac{x}{x+1}(x-1+\log |x|)) is correct because it is the solution to the given differential equation, obtained by finding the integrating factor and applying the initial condition.
163 Medium
163 of 975
Application +4 | -1
Solution of the equation xsin⁡(yx)dy=(ysin⁡yx−x)dxx \sin \left(\frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x is
View answer and explanation
Correct option: C
cos⁡(yx)=log⁡∣kx∣\cos \left(\frac{y}{x}\right)=\log |k x|
Option C (cos⁡(yx)=log⁡∣kx∣\cos \left(\frac{y}{x}\right)=\log |k x|) is correct because it is the solution obtained after solving the given homogeneous differential equation using the substitution y=vxy=vx.
164 Hard
164 of 975
Application +4 | -1
The differential equation of all straight lines touching the circle x2+y2=a2{x^2} + {y^2} = {a^2} is
View answer and explanation
Correct option: B
(y−xdydx)2=a2[1+(dydx)2]{\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = {a^2}\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]
Option B ((y−xdydx)2=a2[1+(dydx)2]{\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = {a^2}\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]) is correct because it is the differential equation representing all straight lines tangent to the circle x2+y2=a2{x^2} + {y^2} = {a^2}.
165 Easy
165 of 975
Application +4 | -1
Which of the following transformation reduce the differential equation dz dx+zxlog⁡z=zx2(log⁡z)2\frac{\mathrm{d} z}{\mathrm{~d} x}+\frac{z}{x} \log z=\frac{z}{x^2}(\log z)^2 into the form du dx+uP(x)=Q(x)\frac{\mathrm{d} u}{\mathrm{~d} x}+u P(x)=Q(x)
View answer and explanation
Correct option: C
u=ezu=e^z
Option B (u=(log⁡z)−1u=(\log z)^{-1}) is correct because it transforms the given differential equation into a linear form suitable for solving.
166 Easy
166 of 975
Understanding +4 | -1
The differential equation of the family of non-vertical lines is .......
View answer and explanation
Correct option: A
d2ydx2=0\frac{d^2 y}{d x^2}=0
Option A (d2ydx2=0\frac{d^2 y}{d x^2}=0) is correct because the general equation of a non-vertical line is y = mx + c, and its second derivative is zero.
167 Easy
167 of 975
Application +4 | -1
If y=asin⁡(5x+c)y = a\sin (5x + c), then
View answer and explanation
Correct option: C
d2ydx2=−25y\frac{{{d^2}y}}{{d{x^2}}} = - 25y
Option C (d2ydx2=−25y\frac{{{d^2}y}}{{d{x^2}}} = - 25y) is correct because the second derivative of the given function equals -25 times the original function.
168 Medium
168 of 975
Understanding +4 | -1
If d2ydx2=0\frac{d^2 y}{d x^2}=0, then
View answer and explanation
Correct option: A
y=ax+by=a x+b
Option A (y=ax+by=a x+b) is correct because integrating the given second derivative equation twice results in a linear equation of the form y = ax + b.
169 Easy
169 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation x3dy+(xy−1)dx=0,x>0,y(12)=3−ex^{3} d y+(x y-1) d x=0, x>0, y\left(\frac{1}{2}\right)=3-e. Then y(1)y(1) is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct. The solution involves solving a first-order linear differential equation and using the given initial condition.
170 Easy
170 of 975
Application +4 | -1
A solution of y=xy′−3y′3y=x y^{\prime}-3 y^{\prime 3} is
View answer and explanation
Correct option: C
y=x−3y=x-3
The correct answer is C (y=x-3) because it satisfies the given Clairaut's equation y=xy'-3y'^3 when y'=1.
171 Easy
171 of 975
Analyzing +4 | -1
If the curve y=y(x)y = y(x) is the solution of the differential equation
2(x2+x5/4)dy−y(x+x1/4)dx=2x9/4dx,  x>02({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})\mathrm{ }dx = 2{x^{9/4}}dx,\,\,x > 0 which passes through the point (1,  1−43log⁡e2)\left( {1,\,\,1 - \frac{4}{3}{{\log }_e}2} \right),then the value of y(16)y(16) is equal to
View answer and explanation
Correct option: C
4(313−83log⁡e3)4\left( {\frac{{31}}{3} - \frac{8}{3}{{\log }_e}3} \right)
Option C (4\left( {\frac{{31}}{3} - \frac{8}{3}{{{\log }_e}3} \right)4\left( {\frac{{31}}{3} - \frac{8}{3}{{{\log }_e}3} \right)) is correct because it is the result of solving the given differential equation and applying the provided initial condition to find y(16).
172 Easy
172 of 975
Application +4 | -1
The value of the constant ' mm ' and ' cc ' for which y=mx+cy=m x+c is a solution of the differential equation D2y−3Dy−4y=−4xD^2 y-3 D y-4 y=-4 x.
View answer and explanation
Correct option: B
is m=1;c=−3/4m=1 ; c=-3 / 4
Option B (m=1 ; c=-3 / 4) is correct because substituting these values into the differential equation satisfies the given condition.
173 Hard
173 of 975
Application +4 | -1
The general solution of differential equation

dydx+yℓnyx=y(ℓny)2x2\frac{dy}{dx}+\frac{y\ell ny}{x}=\frac{y{{(\ell ny)}^{2}}}{{{x}^{2}}}

(where C\mathrm{C} is an arbitrary constant)
View answer and explanation
Correct option: C
1ℓny=12x+Cx\frac{1}{\ell n y}=\frac{1}{2 x}+C x
Option C (1ℓ ny =12x+Cx\frac{{1}}{{\ell \text{{ ny }}}} = \frac{{1}}{{2 x}} + C x) is correct because it is the general solution obtained after solving the given differential equation using a substitution and integrating factor.
174 Easy
174 of 975
Application +4 | -1
If y=cesin⁡−1xy=ce^{\sin ^{-1} x}, then corresponding to this the differential equation is
View answer and explanation
Correct option: A
dydx=y1−x2\frac{d y}{d x}=\frac{y}{\sqrt{1-x^2}}
Option A (dydx=y1−x2\frac{{d y}}{{d x}}=\frac{{y}}{{\sqrt{{1-x^2}}}}) is correct because the derivative of y=cesin⁡−1xy=ce^{\sin^{{-1}}x} satisfies this equation.
175 Medium
175 of 975
Application +4 | -1
If xdy=y(dx+ydy),y(1)=1x\mathrm{ }dy = y\left( {dx + y\mathrm{ }dy} \right),y\left( 1 \right) = 1 and y(x)>0y\left( x \right) > 0, then y(−3)y\left( { - 3} \right) is equal to
View answer and explanation
Correct option: A
3
Option A, 3, is the correct answer because solving the differential equation and applying the initial condition leads to y(-3) = 3.
176 Medium
176 of 975
Application +4 | -1
The solution of the differential equation (x2+y2)dx−5xydy=0,y(1)=0\left(x^2+y^2\right) d x-5 x y d y=0, y(1)=0, is:
View answer and explanation
Correct option: A
∣x2−4y2∣5=x2\left|x^2-4 y^2\right|^5=x^2
Option A (∣x2−4y2∣5=x2\left|x^2-4 y^2\right|^5=x^2) is correct because it is the solution obtained by solving the homogeneous differential equation using the substitution y=Vx and applying the given initial condition.
177 Easy
177 of 975
Application +4 | -1
When x=0,y=1x=0, y=1 the particular solution of the differential equation (1+y2)dx+(1+x2)dy=0\left(1+y^2\right) d x+\left(1+x^2\right) d y=0 is
View answer and explanation
Correct option: B
x+y+xy−1=0x+y+x y-1=0
Option B, x+y+xy−1=0x+y+x y-1=0, is the correct particular solution obtained by solving the differential equation and applying the given initial condition.
178 Easy
178 of 975
Understanding +4 | -1
Statement 1: The degrees of the differential equations dydx+y2=x\frac{d y}{d x}+y^2=x and d2ydx2+y=sin⁡x\frac{d^2 y}{d x^2}+y=\sin x are equal.

Statement 2: The degree of a differential equation, when it is a polynomial equation in derivatives, is the highest positive integral power of the highest order derivative involved in the differential equation, otherwise degree is not defined.
View answer and explanation
Correct option: D
Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 .
Option D (Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1) is correct because both statements are true and statement 2 correctly explains statement 1.
179 Medium
179 of 975
Application +4 | -1
The integral factor of equation (x2+1)dydx+2xy=x2−1\left( {{x^2} + 1} \right)\frac{{dy}}{{dx}} + 2xy = {x^2} - 1 is
View answer and explanation
Correct option: A
x2+1{x^2} + 1
Option C (y2/3−x2/3=c{y^{2/3}} - {x^{2/3}} = c) is correct because it is the solution of the given differential equation after separating variables and integrating.
180 Medium
180 of 975
Application +4 | -1
Solution of 2ysin⁡xdydx=2sin⁡xcos⁡x−y2cos⁡x,x=π2,y=12 y \sin x \frac{d y}{d x}=2 \sin x \cos x-y^{2} \cos x, x=\frac{\pi}{2}, y=1 is given by
View answer and explanation
Correct option: A
y2=sin⁡xy^{2}=\sin x
Option A is correct because solving the differential equation and applying the initial condition results in y2=sin⁡xy^2 = \sin x.
Revision notes and key takeaways

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