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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Mixed difficulty

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481 Medium
481 of 975
Application +4 | -1
Let y=y(x)y=y(x) be the solution of the differential equation (x2−3y2)dx+3xy dy=0, y(1)=1\left(x^{2}-3 y^{2}\right) d x+3 x y \, d y=0, \, y(1)=1. Then 6y2(e)6 y^{2}(e) is equal to
View answer and explanation
Correct option: D
2e22 e^{2}
Option D (2e22e^2) is correct because it involves solving a homogeneous differential equation with an initial condition.
482 Easy
482 of 975
Application +4 | -1
In order to solve the differential equation
xcos⁡xdydx+y(xsin⁡x+cos⁡x)=1x \cos x \frac{d y}{d x}+y(x \sin x+\cos x)=1
the integrating factor is:
View answer and explanation
Correct option: B
xsec⁡xx \sec x
Option B, xsec⁡xx \sec x, is the correct integrating factor because it is derived from the exponential of the integral of P(x) after rewriting the differential equation in standard form.
483 Easy
483 of 975
Application +4 | -1
The solution of (1+xy)ydx+(1−xy)xdy=0(1+x y) y d x+(1-x y) x d y=0 is
View answer and explanation
Correct option: B
log⁡(xy)=1xy+k\log \left(\frac{x}{y}\right)=\frac{1}{x y}+k
Option B (log⁡(xy)=1xy+k\log \left(\frac{x}{y}\right)=\frac{1}{x y}+k) is the correct solution obtained by rearranging the equation, recognizing the exact differential, and integrating both sides.
484 Easy
484 of 975
Application +4 | -1
Solution of ℓn(dy/dx)=4x−2y−2,y(1)=1\ell \mathrm{n}(\mathrm{dy} / \mathrm{dx})=4 \mathrm{x}-2 \mathrm{y}-2, \mathrm{y}(1)=1, is -
View answer and explanation
Correct option: A
2e(2y+2)=e4x+e42 \mathrm{e}^{(2 y+2)}=\mathrm{e}^{4 x}+\mathrm{e}^4
Option A (2e(2y+2)=e4x+e42 \mathrm{e}^{(2 y+2)}=\mathrm{e}^{4 x}+\mathrm{e}^4) is correct because it represents the solution to the given differential equation with the initial condition y(1)=1y(1)=1.
485 Medium
485 of 975
Application +4 | -1
The solution of y′−y=1,y(0)=−1y' - y = 1,y\left( 0 \right) = - 1 is given by y(x)y\left( x \right), which is equal to
View answer and explanation
Correct option: C
−1- 1
Option A (3) is correct because it is the speed of the object in km/h calculated from the change in object position and the given time.
486 Medium
486 of 975
Application +4 | -1
If dydx+y=2e2x\frac{{dy}}{{dx}} + y = 2{e^{2x}}, then yy is equal to
View answer and explanation
Correct option: C
ce−x+23e2xc{e^{ - x}} + \frac{2}{3}{e^{2x}}
Option C (ce−x+23e2xc{e^{ - x}} + \frac{2}{3}{e^{2x}}) is correct because it is the solution to the given first-order linear differential equation obtained using the integrating factor method.
487 Easy
487 of 975
Understanding +4 | -1
The differential equation of all the lines in the xyx y-plane is
View answer and explanation
Correct option: C
d2ydx2=0\frac{d^2 y}{d x^2}=0
Option C (d2ydx2=0\frac{{d^2 y}}{{d x^2}}=0) is correct because the general equation of a line is y = mx + c, and its second derivative with respect to x is always zero.
488 Medium
488 of 975
Application +4 | -1
Let II be the purchase value of an equipment and V(t)V(t) be the value after it has been used for tt years. The value V(t)V(t) depreciates at a rate given by the differential equation dVdt=−k(T−t)\frac{d V}{d t}=-k(T-t), where k>0k>0 is a constant and TT is the total life in years of the equipment. Then the scrap value V(T)V(T) of the equipment is:
View answer and explanation
Correct option: B
I−kT22I-\frac{k T^2}{2}
Option B (I−kT22I-\frac{{k T^2}}{{2}}) is correct because it represents the scrap value V(T) obtained by solving the given differential equation with the initial condition V(0) = I.
489 Easy
489 of 975
Application +4 | -1
Let x=x(y)x = x\left( y \right) be the solution of the differential equation 2yexy2dx+(y2−4xexy2)dy=02y{e^{\frac{x}{{{y^2}}}}}dx + \left( {{y^2} - 4x{e^{\frac{x}{{{y^2}}}}}} \right)dy = 0 such that x(1)=0.x\left( 1 \right) = 0. Then x(e)x\left( e \right) is equal to
View answer and explanation
Correct option: D
−e2log⁡e(2)- {e^2}{\log _e}\left( 2 \right)
Option D (-e²logₑ(2)) is correct because it is the value of x(e) obtained by solving the given differential equation with the initial condition x(1) = 0.
490 Medium
490 of 975
Application +4 | -1
The differential equation of the family of curve y2=4a(x+a){y^2} = 4a\left( {x + a} \right), is
View answer and explanation
Correct option: C
y2(dydx)2+2xydydx−y2=0{y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2xy\frac{{dy}}{{dx}} - {y^2} = 0
Option C (y2(dydx)2+2xydydx−y2=0{y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2xy\frac{{dy}}{{dx}} - {y^2} = 0) is correct because it is the differential equation obtained by eliminating the parameter 'a' from the given family of curves.
491 Medium
491 of 975
Application +4 | -1
The length of the subnormal of the curve x2=4yx^2=4 y at point (−2,1)(-2,1) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the length of the subnormal is calculated using the formula y×dydxy \times \frac{{dy}}{{dx}} evaluated at the given point.
492 Medium
492 of 975
Application +4 | -1
The solution of the differential equation (1+y2)+(x−ettan⁡−1y)dydx=0\left( {1 + {y^2}} \right) + \left( {x - {e^{{\mathop\mathrm{t}\nolimits} {{\tan }^{ - 1}}y}}} \right)\frac{{dy}}{{dx}} = 0 is
View answer and explanation
Correct option: A
2xetan⁡−1y=e2tan⁡−1y+c2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + c
Option A (2xetan⁡−1y=e2tan⁡−1y+c2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + c) is correct because it is the solution obtained by solving the given linear differential equation using the integrating factor method.
493 Hard
493 of 975
Application +4 | -1
If y=axn+1y = a{x^{n + 1}}, then x2d2ydx2{x^2}\frac{{{d^2}y}}{{d{x^2}}} is equal to
View answer and explanation
Correct option: B
n(n+1)yn\left( {n + 1} \right)y
Option B (n(n+1)yn\left( {n + 1} \right)y) is correct because it is the result of calculating the second derivative and multiplying by x2x^2.
494 Medium
494 of 975
Application +4 | -1
d2xdy2\frac{d^2 x}{d y^2} equals to
View answer and explanation
Correct option: B
−(d2ydx2)(dydx)−3-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}
Option B (−(d2ydx2)(dydx)−3-\left(\frac{{d^2 y}}{{d x^2}}\right)\left(\frac{{d y}}{{d x}}\right)^{-3}) is correct because it is the result of applying the chain rule and inverse function differentiation rules.
495 Medium
495 of 975
Application +4 | -1
The solution of dydx=(yx)13\frac{{dy}}{{dx}} = {\left( {\frac{y}{x}} \right)^{\frac{1}{3}}}, is
View answer and explanation
Correct option: C
y2/3−x2/3=C{y^{2/3}} - {x^{2/3}} = C
Option C (y2/3−x2/3=C{y^{2/3}} - {x^{2/3}} = C) is correct because it is the solution obtained by separating variables and integrating the given differential equation.
496 Medium
496 of 975
Application +4 | -1
The solution of the equation dy/dx=cos⁡(x−y)d y / d x=\cos (x-y), where y(3π2)=πy\left(\frac{3 \pi}{2}\right)=\pi is
View answer and explanation
Correct option: B
x+cot⁡(x−y2)=1+3π2x+\cot \left(\frac{x-y}{2}\right)=1+\frac{3 \pi}{2}
Option B (x+cot⁡(x−y2)=1+3π2x+\cot \left(\frac{{x-y}}{{2}}\right)=1+\frac{{3 \pi}}{{2}}) is the correct solution obtained by solving the given differential equation and applying the initial condition.
497 Easy
497 of 975
Analyzing +4 | -1
The order of the differential equation of family of curves y=C1sin⁡−1x+C2cos⁡−1x+C3tan⁡−1x+C4cot⁡−1xy=C_1 \sin ^{-1} x+C_2 \cos ^{-1} x+C_3 \tan ^{-1} x+C_4 \cot ^{-1} x (where C1,C2,C3C_1, C_2, C_3 and C4C_4 are arbitrary constants) is
View answer and explanation
Correct option: C
4
Option D (None of These) is correct because after simplification, the given equation has only two independent arbitrary constants, making the order of the differential equation 2.
498 Medium
498 of 975
Application +4 | -1
The equation of the curve whose subnormal is equal to a constant aa is
View answer and explanation
Correct option: B
y2=2ax+2b{y^2} = 2ax + 2b
Option B (y2=2ax+2b{y^2} = 2ax + 2b) is correct because it is the solution obtained by integrating the differential equation formed from the given condition about the subnormal.
499 Medium
499 of 975
Application +4 | -1
The solution of the D.E ydx−xdy+3x2y2dx=0y d x-x d y+3 x^2 y^2 d x=0 is
View answer and explanation
Correct option: C
x3y+x−cy=0x^3 y+x-c y=0
Option C (x3y+x−cy=0x^3 y+x-c y=0) is correct because it represents the solution obtained by integrating the given differential equation after rewriting it in a suitable form.
500 Medium
500 of 975
Application +4 | -1
dxdy=1+x−y2y\frac{d x}{d y}=\frac{1+x-y^2}{y} and x(1)=1x(1)=1, then 5x(2)5 x(2) is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because solving the differential equation and applying the given condition x(1)=1, we find x(2) = 1, and thus 5x(2) = 5.
501 Easy
501 of 975
Understanding +4 | -1
The differential equation of all parabola whose axis of symmetry is parallel to xx-axis is of order
View answer and explanation
Correct option: B
3
Option B (3) is correct because the general equation of a parabola whose axis of symmetry is parallel to the x-axis has three arbitrary constants, thus the differential equation is of order 3.
502 Easy
502 of 975
Application +4 | -1
If the differential equation of the family of curve given by y=Ax+Be2xy=A x+B e^{2 x}, where AA and BB are arbitrary constants, is of the form (1−2x)ddx(dydx+ly)+k(dydx+ly)=0(1-2 x) \frac{d}{d x}\left(\frac{d y}{d x}+l y\right)+k\left(\frac{d y}{d x}+l y\right)=0, then the ordered pair (k,l)(k, l) is
View answer and explanation
Correct option: A
(2,−2)(2,-2)
Option A, (2,−2)(2,-2), is the correct ordered pair (k,l)(k, l) obtained by finding the differential equation of the given family of curves.
503 Easy
503 of 975
Analyzing +4 | -1
If mm and nn are the order and degree of the differential equation (d2ydx2)5+4(d2ydx2)3d3ydx3+d3ydx3=x2−1,\left(\frac{d^2 y}{d x^2}\right)^5+4 \frac{\left(\frac{d^2 y}{d x^2}\right)^3}{\frac{d^3 y}{d x^3}}+\frac{d^3 y}{d x^3}=x^2-1, then
View answer and explanation
Correct option: B
m=3,n=2m=3, n=2
Option B (m=3,n=2m=3, n=2) is correct because the order of a differential equation is the highest order derivative present, and the degree is the power of the highest order derivative after the equation has been made polynomial in derivatives.
504 Easy
504 of 975
Understanding +4 | -1
Order of the differential equation of the family of all concentric circles centred at (h,k)\left( {h,k} \right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the family of concentric circles is defined by one parameter (the radius), leading to a first-order differential equation.
505 Medium
505 of 975
Application +4 | -1
The tangent at any point PP on the curve y=f(x)y=f(x) meets the yy-axis in B.NB . N is the foot of the perpendicular drawn from PP onto the xx-axis. If the rectangle with OBO B and ON(OO N(O is the origin) as adjacent sides is of constant area and equals a2a^2, then the equation of the curve is
View answer and explanation
Correct option: D
y=cx±a22xy=c x \pm \frac{a^2}{2 x}
Option D (y=cx±a22xy=c x \pm \frac{a^2}{2 x}) is correct because it represents the general solution of the differential equation derived from the given geometric condition.
506 Medium
506 of 975
Application +4 | -1
y=ae−1/x+by=a e^{-1 / x}+b is a solution of dy dx=yx2\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x^2} when
View answer and explanation
Correct option: A
a=1,b=0a=1, b=0
Option A (a=1,b=0a=1, b=0) is correct because it satisfies the given differential equation when substituted into the solution and its derivative.
507 Medium
507 of 975
Application +4 | -1
f(x)+f′(x)=∫02f(t)dtf(x)+f^{\prime}(x)=\int\limits_{0}^2 f(t) d t and f(0)=e−2f(0)=e^{-2}, then the value of f(2)−2f(0)f(2)-2 f(0) is:
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because after solving the differential equation and using the given condition, the value of f(2)-2f(0) is -1.
508 Easy
508 of 975
Application +4 | -1
The curve whose subtangent is nn times the abscissa of the point of contact and passes through (2,3)(2,3) represent
View answer and explanation
Correct option: D
a parabola with vertex origin and axes coincide with xx-axis for n=2n=2
Option D (a parabola with vertex origin and axes coincide with xx-axis for n=2n=2) is correct because the given condition leads to the equation yn=Cxy^n = Cx, which represents a parabola when n=2n=2 with vertex at the origin and axis along the x-axis.
509 Easy
509 of 975
Understanding +4 | -1
Which of the following is a second order of differential equation?
View answer and explanation
Correct option: B
y′y′′+y=sin⁡xy^{\prime} y^{\prime \prime}+y=\sin x
Option B (y′y′′+y=sin⁡xy^{\prime} y^{\prime \prime}+y=\sin x) is correct because it contains a second derivative (y′′y^{\prime \prime}) and no higher-order derivatives.
510 Easy
510 of 975
Application +4 | -1
The area of the region enclosed between the two circles x2+y2=1x^2+y^2=1 and (x−1)2+y2=1(x-1)^2+y^2=1, is
View answer and explanation
Correct option: A
(2π3−34)\left(\frac{2 \pi}{3}-\sqrt{\frac{3}{4}}\right)
Option A is correct because it represents the area of the intersection of two circles, calculated using geometry and trigonometry.
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