Online Test: Differential Equations | Class 12 Mathematics

Question 1 of 10 easy

The order and degree of $(y'')^2+(y')^3+y=0$ are:

2,2

2,3

3,2

1,2

Explanation: Highest derivative is $y''$ (2nd order). Its power is 2. Order=2, Degree=2.

Question 2 of 10 easy

The solution of $\frac{dy}{dx}=e^{x+y}$ is:

$e^x+e^{-y}=C$

$e^{-y}=-e^x+C$

$e^x+e^y=C$

$e^{-x}-e^y=C$

Explanation: Separate: $e^{-y}dy=e^x dx$. Integrate: $-e^{-y}=e^x+C_1$, i.e. $e^{-y}=-e^x+C$.

Question 3 of 10 medium

The degree of $\sqrt{1+(dy/dx)^2}=y$ is:

1

2

3

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Explanation: Square both sides: $1+(dy/dx)^2=y^2$. Degree of highest derivative power = 2.

Question 4 of 10 medium

Integrating factor of $\frac{dy}{dx}-\frac{y}{x}=x^2$ is:

$x$

$1/x$

$-1/x$

$\ln x$

Explanation: $P=-1/x$. IF $=e^{\int(-1/x)dx}=e^{-\ln x}=1/x$.

Question 5 of 10 easy

General solution of $\frac{dy}{dx}+2y=0$ is:

$y=Ce^{2x}$

$y=Ce^{-2x}$

$y=Ce^{x}$

$y=e^{-2x}$

Explanation: Separate: $dy/y=-2dx$. $\ln y=-2x+\ln C$. $y=Ce^{-2x}$.

Question 6 of 10 medium

The family of curves $y=Ae^x+Be^{-x}$ satisfies the DE:

$y''-y=0$

$y''+y=0$

$y'-y=0$

$y''+y'=0$

Explanation: $y'=Ae^x-Be^{-x}$; $y''=Ae^x+Be^{-x}=y$. So $y''-y=0$.

Question 7 of 10 medium

Which DE is homogeneous?

$y'=x+y$

$y'=\frac{x^2+y^2}{xy}$

$y'=x^2+1$

$y'=e^x+y$

Explanation: $\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}$ depends only on $y/x$ — homogeneous.

Question 8 of 10 hard

Solution of $x\frac{dy}{dx}+y=x^2$ with $y(1)=0$ is:

$y=\frac{x^2-1}{3x}$

$y=x^2/3$

$y=x$

$y=\frac{x^3-x}{3}$

Explanation: Linear: $y'+y/x=x$. IF$=x$. $d(xy)/dx=x^2$. $xy=x^3/3+C$. At $x=1,y=0$: $C=-1/3$. $y=x^2/3-1/(3x)=(x^3-1)/(3x)$.

Question 9 of 10 hard

Degree of $(1+y'^2)^{3/2}=ky''$ is:

2

3

1

6

Explanation: Square both sides: $(1+y'^2)^3=k^2(y'')^2$. Expand: max power of $y''$ is 2. Degree=2. Wait: $(1+y'^2)^3$ when expanded has $y'^6$, but $y''$ has power 2. Degree = 2.

Question 10 of 10 medium

$x\frac{dy}{dx}=y(\ln y-\ln x+1)$ is:

Linear

Variable separable

Homogeneous

Exact

Explanation: RHS $=y/x \cdot (\ln(y/x)+1)$ — depends only on $y/x$. Homogeneous.

Online Test — Differential Equations