Exercise 10.9-Additional Problems - Chapter 10 Ordinary Differential Equations Class 12 Samacheer Kalvi Solutions

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Additional Problems
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
The integrating factor of $\frac{d y}{d x}+2 \frac{y}{x}=e^{4 x}$
(a) $\log x$
(b) $x^2$
(c) $e^x$
(d) $x$
Solution:
(b) $x^2$
Hint:
Here $\mathrm{P}=\frac{2}{x} ; \mathrm{Q}=e^{4 x}$
$
\begin{aligned}
\text { I.F. } & =e^{\int P d x} \\
& =e^{\int P d x}=e^{2 \int \frac{1}{x} d x}=e^{2 \log x}=e^{\log x^2}=x^2
\end{aligned}
$
Question 2.
If $\cos \mathrm{x}$ is an integrating factor of the differential equation $\frac{d y}{d x}+P y=Q$ then $P=$
(a) $-\cot x$
(b) $\cot x$
(c) $\tan x$
(d) $-\tan x$
Solution:
(d) $-\tan x$

Question 3.
The integrating factor of $d x+x d y=e^{-y} \sec ^2 y d y$ is
(a) $e^x$
(b) $\mathrm{e}^{-\mathrm{x}}$
(c) $\mathrm{e}^{\mathrm{y}}$
(d) $\mathrm{e}^{-\mathrm{y}}$
Solution:
(c) $\mathrm{e}^{\mathrm{y}}$
Hint:
$
\begin{aligned}
& d x+x d y=e^{-y} \sec ^2 y d y \\
& \frac{d y}{d x}+x=e^{-y} \sec ^2 y \\
& \frac{d y}{d x}+\mathrm{Px}=\mathrm{Q} \\
& \text { Here, } \mathrm{P}=1 ; \mathrm{Q}=e^{-y} \sec ^2 y \\
& \text { I.F. }=e^{\int \mathrm{Pdx}}=e^{\int d y}=e^y
\end{aligned}
$
Question 4.
Integrating factor of $\frac{d y}{d x}+\frac{1}{x \log x} \cdot y=\frac{2}{x^2}$ is .....
(a) $\mathrm{e}^{\mathrm{x}}$
(b) $\log \mathrm{x}$
(c) $\frac{1}{x}$
(d) $\mathrm{e}^{-\mathrm{x}}$
Solution:
(b) $\log x$
Hint:
Here $\mathrm{P}=\frac{1}{x \log x} ; \mathrm{Q}=\frac{2}{x^2}$
$
\text { I.F. }=e^{\int P d x}=e^{\frac{1}{x \log x}}=e^{\log (\log x)}=\log x
$

Question 5.
Solution of $\frac{d y}{d x}+m x=0$, where $\mathrm{m}<0$ is .....
(a) $x=c e^{m y}$
(b) $x=c e^{-m y}$
(c) $x=m y+c$
(d) $x=c$
Solution:
(b) $
x=c e^{-m y}
$
Hint:
$
\begin{aligned}
& \frac{d y}{d x}+m x=0 \\
& \frac{d x}{d y}=-m x \Rightarrow \frac{d x}{x}=-m d y \\
& \text { Integrating both sides, } \\
& \log x=-m y+\log c \Rightarrow \log x-\log c=-m y \\
& \frac{x}{c}=e^{-m y} \Rightarrow x=c e^{-m y}
\end{aligned}
$
Question 6.
$\mathrm{y}=\mathrm{cx}-\mathrm{c}^2$ is the general solution of the differential equation .......
(a) $\left(y^{\prime}\right)^2-x y^{\prime}+y=0$
(b) $y^{\prime \prime}=0$
(c) $y^{\prime}=c$
$(d)\left(y^{\prime}\right)^2+x y^{\prime}+y=0$
Solution:
(a) $\left(y^{\prime}\right)^2-x y^{\prime}+y=0$
Hint:
$
\begin{aligned}
y & =c x-c^2 \\
y^{\prime \prime} & =c \\
\therefore y & =y^{\prime} x-\left(y^{\prime}\right)^2 \Rightarrow\left(y^{\prime}\right)^2-x y^{\prime}+y=0
\end{aligned}
$

Question 7.
The differential equation of all non-vertical lines in a plane is ......
(a) $\frac{d y}{d x}=0$
(b) $\frac{d^2 y}{d x^2}=0$
(c) $\frac{d y}{d x}=m$
(d) $\frac{d^2 y}{d x^2}=m$
Solution:
(b) $
\frac{d^2 y}{d x^2}=0
$
Hint:
The equation of the straight line is $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
$
\Rightarrow \frac{d y}{d x}=m \Rightarrow \frac{d^2 y}{d x^2}=0
$
Question 8.
The differential equation of all circles with centre at the origin is
(a) $x d y+y d x=0$
(b) $x \mathrm{dy}-\mathrm{y} d x=0$
(c) $x d x+y d y=0$
(d) $x \mathrm{dx}-\mathrm{y} d \mathrm{y}=0$
Solution:
(c) $x d x+y d y=0$
Hint:
The equation of family of circle with the centre at the origin is $x^2+y^2=a^2$
$
2 x+2 y \frac{d y}{d x}=0 \Rightarrow x d x+y d y=0
$

The equation of family of circle with the centre at the origin is $x^2+y^2=a^2$
$
2 x+2 y \frac{d y}{d x}=0 \Rightarrow x d x+y d y=0
$
Question 9.
The differential equation of the family of lines $\mathrm{y}=\mathrm{mx}$ is
(a) $\frac{d y}{d x}=m$
(b) $y d x-x d y=0$
(c) $\frac{d^2 y}{d x^2}=0$
(d) $y d x+x d y=0$
Solution:
(d) 6
Hint:
$
\begin{aligned}
& \text { Given, } \begin{aligned}
y & =m x+c \Rightarrow \frac{d y}{d x}=m \\
\therefore y & =\frac{d y}{d x}=x \\
y d x & =x d y \Rightarrow y d x-x d y=0
\end{aligned}
\end{aligned}
$
Question 10.
The degree of the differential equation $\sqrt{1+\left(\frac{d y}{d x}\right)^{1 / 3}}=\frac{d^2 y}{d x^2}$
(a) 1
(b) 2
(c) 3
(d) 6
Solution:
(d) 6

Hint:
$
\sqrt{1+\left(\frac{d y}{d x}\right)^{1 / 3}}=\frac{d^2 y}{d x^2} .
$
Squaring on both sides
$
\begin{aligned}
1+\left(\frac{d y}{d x}\right)^{1 / 3} & =\left(\frac{d^2 y}{d x^2}\right)^2 \\
\left(\frac{d y}{d x}\right)^{1 / 3} & =\left(\frac{d^2 y}{d x^2}\right)^2-1
\end{aligned}
$
Raising to the power 3 on both sides
$
\frac{d y}{d x}=\left(\left(\frac{d^2 y}{d x^2}\right)-1\right)^3
$
Order $=2$, degree $=6$
Question 11.
The degree of the differential equation $c=\frac{\left[1+\left(\frac{d y}{d x}\right)^3\right]^{2 / 3}}{\frac{d^3 y}{d x^3}}$ where $c$ is a constant is
(a) 1
(b) 3
(c) $-2$
(d) 2
Solution:
(b) 3

Hint:
$
\begin{array}{r}
c=\frac{\left[1+\left(\frac{d y}{d x}\right)^3\right]^{2 / 3}}{\frac{d^3 y}{d x^3}} \\
c\left(\frac{d^3 y}{d x^3}\right)=\left[1+\left(\frac{d y}{d x}\right)^3\right]^{\frac{2}{3}}
\end{array}
$
Raising to the power 3 on both sides
$
c^3\left(\frac{d^3 y}{d x^3}\right)^3=\left[1+\left(\frac{d y}{d x}\right)^3\right]^2
$
Order $=3$, degree $=3$

Question 12.
The amount present in a radio active element disintegrates at a rate proportional to its amount. The differential equation corresponding to the above statement is ( $\mathrm{k}$ is negative)
(a) $\frac{d p}{d t}=\frac{k}{p}$
(b) $\frac{d p}{d t}=k t$
(c) $\frac{d p}{d t}=k p$
(d) $\frac{d p}{d t}=-k t$.
Solution:
(c) $\frac{d p}{d t}=k p$
Hint:
Let $\mathrm{p}$ be the amount present in a radio active element
$
\begin{aligned}
\frac{d p}{d t} & =-(-k p) \quad[\because k \text { is negative }] \\
\Rightarrow \frac{d p}{d t} & =k p
\end{aligned}
$
Question 13.
On putting $y=v x$, the homogeneous differential equation $x^2 d y+y(x+y) d x=0$ becomes ......
(a) $x d v+\left(2 v+v^2\right) d x=0$
(b) $v d x+\left(2 x+x^2\right) d v=0$
(c) $v^2 d x-\left(x+x^2\right) d v=0$
(d) $v d v+\left(2 x+x^2\right) d x=0$
Solution:
(a) $x d v+\left(2 v+v^2\right) d x=0$

Hint:
Given, $x^2 d y+y(x+y) d x=0$
$
\begin{aligned}
\Rightarrow x^2 d y & =-y(x+y) d x \\
\frac{d y}{d x} & =\frac{-y x-y^2}{x^2}=\frac{-y}{x} \frac{-y^2}{x^2}
\end{aligned}
$
Putting $y=v x \Rightarrow \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}$
Substituting in $(1) \Rightarrow v+x \cdot \frac{d v}{d x}=-v-v^2$
$
\begin{aligned}
x \cdot \frac{d v}{d x} & =-v^2-2 v=-\left(v^2+2 v\right) \\
x \cdot d v & =-\left(v^2+2 v\right) \cdot d x \\
x d v+\left(2 v+v^2\right) d x & =0
\end{aligned}
$