Exercise 10.4-Additional Problems - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Verify that the function $\mathrm{y}=\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}$ is a solution of the differential equation $\cos \frac{d y}{d x}+\mathrm{y} \sin \mathrm{x}=$ b. $\mathrm{dx}$
Solution:
The given function is $y=a \cos x+b \sin x$
Differentiating both sides with respect to $\mathrm{x}$, we have
$
\frac{d y}{d x}=-a \sin x+b \cos x
$
Putting values of $\frac{d y}{d x}$ and $\mathrm{y}$ in the given differential equation, we have
L.H.S. $=\cos x(-a \sin x+b \cos x)+\{a \cos x+b \sin x) \sin x$
$=-a \sin x \cos x+b \cos 2 x+a \sin x \cos x+b \sin 2 x=b(\cos 2 x+\sin 2 x)$
$=\mathrm{b} \times 1=\mathrm{b}=$ R.H.S
Thus, $\mathrm{y}=\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}$ is a solution of differential equation
$
\cos x \frac{d y}{d x}+y \sin x=b
$

Question 2.
Verify that the function $y=4 \sin 3 x$ is a solution of the differential equation $\frac{d^2 y}{d x^2}+9 y=0$
Solution:
The given function is $y=4 \sin 3 x$
Differentiating both sides with respect to $\mathrm{x}$, we have
$
\frac{d y}{d x}=4(3 \cos 3 x)=12 \cos 3 x
$
Again, differentiating both sides with respect to $\mathrm{x}$, we have
$
\frac{d^2 y}{d x^2}=12(-3 \sin 3 x)=-36 \sin 3 x
$
Putting values of $\frac{d^2 y}{d x^2}$ and $y$ in the given differential equation, we have
L.H.S. $=-36 \sin 3 x+9(4 \sin 3 x)=-36 \sin 3 x+36 \sin 3 x=0=$ R.H.S.
Thus, $\mathrm{y}=4 \sin 3 \mathrm{x}$ is a solution of differential equation $\frac{d^2 y}{d x^2}+9 y=0$

Question 3.
Verify that the function $\mathrm{y}=\mathrm{ax} \mathrm{ax}^2+\mathrm{bx}+\mathrm{c}$ is a solution of the differential equation $\frac{d^2 y}{d x^2}=2 a$.

Solution:
The given function is $y=a x^2+b x+c$
Differentiating both sides with respect to $\mathrm{x}$, we have
$
\frac{d y}{d x}=2 a x+b
$
Again differentiating both sides with respect to $\mathrm{x}$, we have
$
\frac{d^2 y}{d x^2}=2 a
$
Which is the given differential equation.
Thus, $\mathrm{y}=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}$ is a solution of differential equation $\frac{d^2 y}{d x^2}=2 \mathrm{a}$.
Question 4.
Verify that the function $y=\mathrm{e}^{-3 \mathrm{x}}$ is a solution of the differential equation
$
\frac{d y}{d x}=-3 e^{-3 x}
$
Solution:
The given function is $y=e^{-3 x}$
Differentiating both sides with respect to $\mathrm{x}$, we have
$
\frac{d y}{d x}=-3 e^{-3 x}
$

Again, differentiating both sides with respect to $\mathrm{x}$, we have
$
\frac{d^2 y}{d x^2}=-3\left(-3 e^{-3 x}\right)=9 e^{-3 x}
$
Putting values of $\frac{d^2 y}{d x^2}, \frac{d y}{d x}$ and $y$ in the given differential equation, we have L.H.S. $=9 e^{-3 x}-3 e^{-3 x}-6 e^{-3 x}=9 e^{-3 x}-9 e^{-3 x}=0=$ R.H.S.
Thus, $y=e^{-3 x}$ is a solution of differential equation $\frac{d^2 y}{d x^2}+\frac{d y}{d x}-6 y=0$.