Exercise 10.1-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 order and degree of the following differential equations:
(i) $\frac{d y}{d x}+y=x^2$
Solution:
Order $=1$,
Degree $=1$
$\left[\because\right.$ we have $\left.\left(\frac{d y}{d x}\right)^{\prime}\right]$
(ii) $y^{\prime}+y^2+y^3=0$
Solution:
Order $=1$,
Degree $=1$
$\left[\because\right.$ we have $\left.\left(\frac{d y}{d x}\right)^{\prime}\right]$
(iii) $y^{\prime \prime}+3 y^{\prime 2}+y^3=0$
Solution:
We have $\left(\frac{d^2 y}{d x^2}\right)^{\prime} ;$ So, order $=2 ;$ Degree $=1$

(iv) $\frac{d^2 y}{d x^2}+x=\sqrt{y+\frac{d y}{d x}}$
Solution:
Squaring both sides, $\left[\frac{d^2 y}{d x^2}+x\right]^2=y+\frac{d y}{d x}$
We have $\left(\frac{d^2 y}{d x^2}\right)^2$
$\therefore$ Order $=2$
Degree $=2$
(v) $\frac{d^2 y}{d x^2}-y+\left(\frac{d y}{d x}+\frac{d^3 y}{d x^3}\right)^{\frac{3}{2}}=0$
Solution:
$
\frac{d^2 y}{d x^2}-y=-\left(\frac{d y}{d x}+\frac{d^3 y}{d x^3}\right)^{\frac{3}{2}}
$
Squaring both sides, we get, $\left(\frac{d^2 y}{d x^2}-y\right)^2=\left(\frac{d y}{d x}+\frac{d^3 y}{d x^3}\right)^3$
We have $\left(\frac{d^3 y}{d x^3}\right)^3 \quad$ Order $=3 ;$ Degree $=3$

(vi) $y^{\prime \prime}=\left(y-y^{\prime 3}\right)^{2 / 3}$
Solution:
$
y^{\prime \prime}=\left(y-y^{\prime 3}\right)^{2 / 3}
$
Raising this equation to the power three, we get, $\left(y^{\prime \prime}\right)^3=\left(y-y^{\prime 3}\right)^2$
We have $\left(\frac{d^2 y}{d x^2}\right)^3$. Order $=2 ;$ Degree $=3$.
(vii) $y^{\prime}+\left(y^{\prime \prime}\right)^2=\left(x+y^{\prime \prime}\right)^2$
Solution:
$
\begin{aligned}
y^{\prime}+\left(y^{\prime \prime}\right)^2 & =x^2+y^{\prime \prime 2}+2 x y^{\prime \prime} \\
\Rightarrow x^2+2 x y^{\prime \prime}-y^{\prime} & =0
\end{aligned}
$
We have $\left(\frac{d^2 y}{d x^2}\right)^{\prime} \Rightarrow$ Order $=2 ;$ Degree $=1$
(viii) $y^{\prime}+\left(y^{\prime \prime}\right)^2=x\left(x+y^{\prime \prime}\right)^2$
Solution:
$
\begin{aligned}
y^{\prime}+\left(y^{\prime \prime}\right)^2 & =x\left[x^2+\left(y^{\prime \prime}\right)^2+2 x y^{\prime \prime}\right] \\
& =y^{\prime}+\left(y^{\prime \prime}\right)^2(1-x)=x^3+2 x^2 y^{\prime \prime}
\end{aligned}
$
We have $\left(y^{\prime \prime}\right)^2 . \Rightarrow$ Order $=2 ;$ Degree $=2$

(ix) $\left(\frac{d y}{d x}\right)^2+x=\left(\frac{d x}{d y}\right)+x^2$
Solution:
$
\begin{aligned}
& \left(\frac{d y}{d x}\right)^2+x=\frac{1}{\frac{d y}{d x}}+x^2 \\
& \left(\frac{d y}{d x}\right)^3+x \frac{d y}{d x}=1+x^2 \frac{d y}{d x} ; \text { we have }\left(\frac{d y}{d x}\right)^3 . \\
& \therefore \quad \text { Order }=1 \text {; Degree }=3
\end{aligned}
$
(x) $\sin x(d x+d y)=\cos x(d x-d y)$
Solution:
Order $=1$;
Degree $=1$