Exercise 10.3-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.
Find the differential equation of the family of straight lines $\mathrm{y}=\mathrm{mx}+\frac{a}{m}$ when (i) $m$ is the parameter,
(ii) a is the parameter,
(iii) a, $m$ both are parameters.
Solution:
(i) $y=m x+\frac{a}{m}$
$
\therefore y^{\prime}=m \Rightarrow y=x y^{\prime}+\frac{a}{y^{\prime}} \Rightarrow y y^{\prime}=x\left(y^{\prime}\right)^2+a
$
(ii) $y=m x+\frac{a}{m} ; \frac{d y}{d x}=m \quad$ i.e., $y^{\prime}=m$
(iii) $y=m x+\frac{a}{m} ; y^{\prime}=m ; \quad y^{\prime \prime}=0$

Question 2.
Find the differential equation that will represent family of all circles having centres on the $x$-axis and the radius is unity.
Solution:
Equation of a circle with centre on $\mathrm{x}$-axis and radius 1 unit is
$(x-a)^2+y^2=1$
Differentiating with respect to $\mathrm{x}$
$
\begin{aligned}
& 2(x-a)+2 y^{\prime}-0 \\
& \Rightarrow 2(x-a)=-2 y y^{\prime}
\end{aligned}
$
(or) $\mathrm{x}-\mathrm{a}=-\mathrm{yy} \mathrm{y}^{\prime} \ldots \ldots$ (2)
Substituting (2) in (1), we get,
$
\begin{aligned}
\left(-y y^{\prime}\right)^2+y^2 & =1 \\
\text { (i.e.,) } y^2\left(y^{\prime}\right)^2+y^2 & =1 \Rightarrow y^2\left[1+\left(y^{\prime}\right)^2\right]=1
\end{aligned}
$

Question 3.
From the differential equation from the following equations.
(i) $y=e^{2 x}(A+B x)$
Solution:
$\mathrm{ye}^{-2 \mathrm{x}}=\mathrm{A}+\mathrm{Bx}$
Since the above equation contains two arbitrary constants, differentiating twice,
we get $y^{\prime} e^{-2 x}-2 y e^{-2 x}=\mathrm{B}$
$
\begin{array}{lll}
\left(y^{\prime \prime} e^{-2 x}-2 y^{\prime} e^{-2 x}\right) & -2\left(y^{\prime} e^{-2 x}-2 y e^{-2 x}\right)=0 \\
e^{-2 x}\left(y^{\prime \prime}-4 y^{\prime}+4 y\right) & =0 & {\left[\because e^{-2 x} \neq 0\right]}
\end{array}
$
$y^{\prime \prime}-4 y^{\prime}+4 y=0$ is the required differential equation.
(ii) $y=e^x(A \cos 3 x+B \sin 3 x)$
Solution:

$
y e^{-x}=\mathrm{A} \cos 3 x+\mathrm{B} \sin 3 x
$
We have to differentiate twice to eliminate two arbitrary constants
$
\begin{array}{rlr}
y^{\prime} e^{-x}-y e^{-x} & =-3 \mathrm{~A} \sin 3 x+3 \mathrm{~B} \cos 3 x \\
y^{\prime \prime} e^{-x}-y^{\prime} e^{-x}-y^{\prime} e^{-x}+y e^{-x} & =-9(\mathrm{~A} \cos 3 x+\mathrm{B} \sin 3 x) \\
\text { i.e., } \quad e^{-x}\left(y^{\prime \prime}-2 y^{\prime}+y\right) & =-9 y e^{-x} & \\
\Rightarrow y^{\prime \prime}-2 y^{\prime}+10 y & =0 & {\left[\because e^{-x} \neq 0\right]}
\end{array}
$
(iii) $\mathrm{Ax}^2+\mathrm{By}^2=1$
Solution:
Differentiating, $\quad 2 \mathrm{~A} x+2 \mathrm{~B} y y^{\prime}=0$ i.e., $\mathrm{A} x+\mathrm{B} y y^{\prime}=0$
Differentiating again, $\mathrm{A}+\mathrm{B}\left(y y^{\prime \prime}+y^{\prime 2}\right)=0$
Eliminating A and B between (1), (2) and (3) we get
$
\left|\begin{array}{ccc}
x^2 & y^2 & -1 \\
x & y y^{\prime} & 0 \\
1 & y y^{\prime \prime}+y^{\prime 2} & 0
\end{array}\right|=0 \Rightarrow\left(y y^{\prime \prime}+y^{\prime 2}\right) x-y y^{\prime}=0
$
(iv) $y^2=4 a(x-a)$
Solution:
$
y^2=4 a(x-a)
$
Differentiating, $2 y y^{\prime}=4 a$
Eliminating $a$ between (1) and (2) we get
$
\begin{aligned}
y^2 & =2 y y^{\prime}\left(x-\frac{y y^{\prime}}{2}\right) \\
\Rightarrow\left(y y^{\prime}\right)^2-2 x y y^{\prime}+y^2 & =0
\end{aligned}
$