Exercise 10.1-Additional Problems - Chapter 10 Ordinary Differential Equations Class 12 Samacheer Kalvi Solutions

You can Download the Exercise 10.1-Additional Problems - Chapter 10 Ordinary Differential Equations Class 12 Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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 $=1$