Exercise 10.5 - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $10.5$
Question 1.
If $F$ is the constant force generated by the motor of an automobile of mass $M$, its velocity $V$ is given by $\mathrm{M} \frac{d \mathbf{V}}{d t}=\mathrm{F}-\mathrm{kV}$, where $\mathrm{A}$ is a constant. Express $\mathrm{V}$ in terms of $\mathrm{t}$ given that $\mathrm{V}=0$ when $\mathrm{t}=0$.
Solution:
Given equation $\mathrm{M} \frac{d \mathrm{~V}}{d t}=\mathrm{F}-k \quad(\because \mathrm{F}$ and $k$ are constant)
$
\begin{aligned}
\mathrm{M} \frac{d \mathrm{~V}}{d t} & =k\left(\frac{\mathrm{F}}{k}-\mathrm{V}\right) \\
\int \frac{d \mathrm{~V}}{\left(\frac{\mathrm{F}}{k}-\mathrm{V}\right)} & =\frac{k}{\mathrm{M}} \int d t \\
-\log \left(\frac{\mathrm{F}}{k}-\mathrm{V}\right) & =\frac{k}{\mathrm{M}} t+c
\end{aligned}
$
Given $\mathrm{V}=0$ and $t=0$
$
\Rightarrow-\log \frac{\mathrm{F}}{k}=c
$
Substituting in (1)
$
\begin{aligned}
& -\log \left(\frac{\mathrm{F}}{k}-\mathrm{V}\right)=\frac{k t}{\mathrm{M}}-\log \left(\frac{\mathrm{F}}{k}\right) \\
& \log \left(\frac{\mathrm{F}}{k}\right)-\log \left(\frac{\mathrm{F}-\mathrm{V} k}{k}\right)=\frac{k t}{\mathrm{M}} \\
& \log \left(\frac{\mathrm{F} / k}{\frac{\mathrm{F}-\mathrm{V} k}{k}}\right)=\frac{k t}{\mathrm{M}} \\
& \left(\frac{\mathrm{F}}{\mathrm{F}-\mathrm{V} k}\right)=e^{\frac{k t}{M}} \\
& \mathrm{~F}=(\mathrm{F}-\mathrm{V} k) e^{\frac{k t}{\mathrm{M}}} \\
&
\end{aligned}
$

Question 2.
The velocity $\mathrm{v}$, of a parachute falling vertically satisfies the equation,
$
v \frac{d v}{d x}=g\left(1-\frac{v^2}{k^2}\right)
$
where g and $\mathrm{k}$ are constants. If $\mathrm{v}$ and $\mathrm{x}$ are both initially zero, find $\mathrm{v}$ in terms of $\mathrm{x}$.
Solution:
Given $v \frac{d v}{d x}=g\left(1-\frac{v^2}{k^2}\right)$
$
\begin{aligned}
v \frac{d v}{d x} & =g\left(\frac{k^2-v^2}{k^2}\right) \\
\int \frac{v d v}{k^2-v^2} & =\frac{g}{k^2} \int d x \\
-\frac{1}{2} \log \left(k^2-v^2\right) & =\frac{g}{k^2} x+c \\
\text { given } v=0 \text { and } x & =0 \\
-\frac{1}{2} \log k^2 & =c \Rightarrow \log \frac{1}{k}=c
\end{aligned}
$
Substitute in (1)
$
\begin{aligned}
& -\frac{1}{2} \log \left(k^2-v^2\right)=\frac{g}{k^2} x+\log \frac{1}{k} \\
& \log \left(\frac{1}{\sqrt{k^2-v^2}}\right)-\log \left(\frac{1}{k}\right)=\frac{g}{k^2} x \\
& \log \left(\frac{k}{\sqrt{k^2-v^2}}\right)=\frac{g x}{k^2} \\
& \frac{k}{\sqrt{k^2-v^2}}=e^{\frac{g x}{k^2}} \\
& k e^{-g x / k^2}=\sqrt{k^2-v^2} \\
&
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
k^2 e^{-2 g x} / k^2 & =k^2-v^2 \\
v^2 & =k^2\left(1-e^{-2 g x / k^2}\right)
\end{aligned}
$

Question 3.
Find the equation of the curve whose slope is $\frac{y-1}{x^2+x}$ and which passes through the point $(1,0)$.

Solution:
Given that the slope is $\frac{y-1}{x^2+x}$
$
\begin{aligned}
\text { i.e., } \frac{d y}{d x} & =\frac{y-1}{x^2+x} \\
\int \frac{d y}{y-1} & =\int \frac{d x}{x^2+x} \\
\log (y-1) & =\int \frac{(x+1)-x}{x(x+1)} d x \\
\log (y-1) & =\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x \\
\log (y-1) & =\log x-\log (x+1)+\log c \\
\log (y-1) & =\log \left(\frac{c x}{x+1}\right) \\
y-1 & =\frac{c x}{x+1}
\end{aligned}
$
This passes through $(1,0)$
$
\Rightarrow-1=\frac{c}{2} \Rightarrow c=-2
$
$\therefore$ Equation of the curve is $y-1=\frac{-2 x}{x+1}$
$
\begin{aligned}
& y=1-\frac{2 x}{x+1}=\frac{x+1-2 x}{x+1} \\
& y=\frac{1-x}{x+1}
\end{aligned}
$

Question 4.
Solve the following differential equations:
(i) $\frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}$
Solution:
$
\begin{aligned}
\frac{d y}{d x} & =\sqrt{\frac{1-y^2}{1-x^2}} \\
\frac{d y}{d x} & =\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}
\end{aligned}
$
Separating the variables
$
\begin{aligned}
\int \frac{d y}{\sqrt{1-y^2}} & =\int \frac{d x}{\sqrt{1-x^2}} \\
\sin ^{-1} y & =\sin ^{-1} x+c
\end{aligned}
$

(ii) $y d x+\left(1+x^2\right) \tan ^{-1} x d y=0$
Solution:
$
\begin{aligned}
& y d x+\left(1+x^2\right) \tan ^{-1} x d y=0 \\
& \left(1+x^2\right) \tan ^{-1} x d y=-y d x
\end{aligned}
$
Separating the variables
$
\begin{aligned}
& \int \frac{d y}{y}=-\int \frac{d x}{\left(1+x^2\right) \tan ^{-1} x} \\
& \log \mathrm{y}=-\log \left(\tan ^{-1} \mathrm{x}\right)+\log \mathrm{c} \\
& \log \mathrm{y}+\log \left(\tan ^{-1} \mathrm{x}\right)=\log \mathrm{c} \\
& \log \left(\mathrm{y} \tan ^{-1} \mathrm{x}\right)=\log \mathrm{c} \\
& \mathrm{y} \tan ^{-1} \mathrm{x}=\mathrm{c}
\end{aligned}
$
(iii) $\sin \frac{d y}{d x}=\mathrm{a}, \mathrm{y}(0)=1$
Solution:
$
\begin{aligned}
\sin \left(\frac{d y}{d x}\right) & =a, y(0)=1 \\
\frac{d y}{d x} & =\sin ^{-1} a \\
\int d y & =\left(\sin ^{-1} a\right) \int d x \\
y & =x\left(\sin ^{-1} a\right)+c
\end{aligned}
$
Given $y(0)=1$
i.e., When $x=0, y=1$
$
\therefore 1=0\left(\sin ^{-1} \mathrm{a}\right)+\mathrm{c} \Rightarrow \mathrm{c}=1
$

$\begin{aligned}
& \therefore \text { The solution is } \mathrm{y}=\mathrm{xsin}^{-1} \mathrm{a}+1 \\
& y-1=x \sin ^{-1} a \\
& \frac{y-1}{x}=\sin ^{-1} a \\
& \sin \left(\frac{y-1}{x}\right)=a
\end{aligned}$

(iv) $\frac{d y}{d x}=e^{x+y}+x^3 e^y$
Solution:
$
\begin{aligned}
\frac{d y}{d x} & =e^{x+y}+x^3 e^y \\
\frac{d y}{d x} & =e^x \cdot e^y+x^3 e^y \\
\frac{d y}{d x} & =e^y\left(e^x+x^3\right) \\
\int e^{-y} d y & =\int\left(e^x+x^3\right) d x \\
-e^{-y} & =e^x+\frac{x^4}{4}+c \\
e^x+e^{-y}+\frac{x^4}{4} & =-c \Rightarrow e^x+e^{-y}+\frac{x^4}{4}=c
\end{aligned}
$

(v) $\left(e^{\mathrm{y}}+1\right) \cos \mathrm{x} d \mathrm{x}+\mathrm{e}^{\mathrm{y}} \sin \mathrm{x} d \mathrm{y}=0$
Solution:
$
\begin{aligned}
& \left(\mathrm{e}^{\mathrm{y}}+1\right) \cos \mathrm{xdx}+\mathrm{e}^{\mathrm{y}} \sin \mathrm{x} d \mathrm{y}=0 \\
& \mathrm{e}^{\mathrm{y}} \sin \mathrm{x} d \mathrm{y}=-\left(\mathrm{e}^{\mathrm{y}}+1\right) \cos \mathrm{x} d \mathrm{x} \\
& \int \frac{e^y d y}{e^y+1}=-\int \frac{\cos x d x}{\sin x} \\
& \log \left(\mathrm{e}^{\mathrm{y}}+1\right)=-\log \sin \mathrm{x}+\log \mathrm{c} \\
& \log \left[\left(\mathrm{e}^{\mathrm{y}}+1\right)+\log \sin \mathrm{x}=\log \mathrm{c}\right. \\
& \left.\log \left(\mathrm{e}^{\mathrm{y}}+1\right) \sin \mathrm{x}\right]=\log \mathrm{c} \\
& \left(\mathrm{e}^{\mathrm{y}}+1\right) \sin \mathrm{x}=\mathrm{c}
\end{aligned}
$

(vi) $(y d x-x d y) \cot \left(\frac{x}{y}\right)=n y^2 d x$
Solution:
$
(y d x-x d y) \cot \left(\frac{x}{y}\right)=n y^2 d x
$
Dividing throughout by ' $y^2$
$
\begin{aligned}
\left(\frac{y d x-x d y}{y^2}\right) \cot \left(\frac{x}{y}\right) & =n d x \\
d\left(\frac{x}{y}\right) \cdot \cot \left(\frac{x}{y}\right) & =n d x \\
\int \cot \left(\frac{x}{y}\right) \cdot d\left(\frac{x}{y}\right) & =n \int d x \\
\log \sin \left(\frac{x}{y}\right) & =n x+c \\
\sin \left(\frac{x}{y}\right) & =e^{n x+c}
\end{aligned}
$

(vii) $\frac{d y}{d x}-x \sqrt{25-x^2}=0$

Solution:

$\text { put } \begin{aligned}
t & =25-x^2 \\
d t & =-2 x d x \\
\frac{d t}{-2} & =x d x
\end{aligned}$
$
\begin{aligned}
\frac{d y}{d x}-x \sqrt{25-x^2} & =0 \\
\frac{d y}{d x} & =x \sqrt{25-x^2} \\
\int d y & =\int x \sqrt{25-x^2} d x \Rightarrow \int d y=-\frac{1}{2} \int \sqrt{t} d t \\
y & =-\frac{1}{2} \cdot \frac{2}{3} t^{3 / 2}+c \\
3 y & =-t^{3 / 2}+3 c \\
\therefore 3 y & =-\left(25-x^2\right)^{3 / 2}+3 c
\end{aligned}
$

(viii) $\mathrm{x} \cos \mathrm{y} d \mathrm{y}=\mathrm{e}^{\mathrm{x}}(\mathrm{x} \log \mathrm{x}+1) \mathrm{dx}$
Solution:
$
\begin{aligned}
x \cos y d y & =e^x(x \log x+1) d x \\
\cos y d y & =e^x\left(\frac{x \log x+1}{x}\right) d x \\
\cos y d y & =e^x\left(\log x+\frac{1}{x}\right) d x \\
\int \cos y d y & =\int e^x \log x d x+\int e^x \cdot \frac{1}{x} d x
\end{aligned}
$
Let us integrate $e^x \log x$ by using integration by parts
$
\begin{aligned}
\int e^x \log x d x & \\
\operatorname{Let} u & =\log x \\
d u & =\frac{1}{x} d x \\
\int u d v & =u v-\int v d u \\
\int e^x \log x d x & =e^x \log x-\int e^x \cdot \frac{1}{x} d x \\
\text { Substitute in } & (1) \\
\int \cos y d y & =\int e^x \\
\int \sin y & =e^x \log x d x+\int e^x d x \\
\cdot \quad \sin y & =e^x \log x+c
\end{aligned}
$

(ix) $
\tan y \frac{d y}{d x}=\cos (x+y)+\cos (x-y)
$
Solution:
$
\begin{aligned}
& \tan y \frac{d y}{d x}=\cos (x+y)+\cos (x-y) \\
& \tan y \frac{d y}{d x}=\cos x \cos y-\sin x \sin y+\cos x \cos y+\sin x \sin y \\
& \tan y \frac{d y}{d x}=2 \cos x \cos y
\end{aligned}
$
Separating the variables
$
\begin{aligned}
\int \frac{\tan y}{\cos y} d y & =2 \int \cos x d x \Rightarrow \int \sec y \tan y d y=2 \int \cos x d x \\
\sec y & =2 \sin x+c
\end{aligned}
$

$\text { (x) } \frac{d y}{d x}=\tan ^2(x+y)$

Solution:
$
\frac{d y}{d x}=\tan ^2(\mathrm{x}+\mathrm{y})
$
Let $u=x+y$
Differentiating with respect to ' $x$ '
$
\begin{aligned}
& \frac{d u}{d x}=1+\frac{d y}{d x} \\
& \frac{d u}{d x}-1=\frac{d y}{d x} \\
& \text { Now, } \frac{d y}{d x}=\tan ^2(x+y) \\
& \frac{d u}{d x}-1=\tan ^2 u \\
& \frac{d u}{d x}=1+\tan ^2 u \\
& \frac{d u}{d x}=\sec ^2 u \\
& \frac{d u}{\sec ^2 u}=d x \\
& \int \cos ^2 u d u=\int d x \\
& \int\left(\frac{1+\cos 2 u}{2}\right) d u=x+c \\
& \frac{1}{2}\left[u+\frac{\sin 2 u}{2}\right]=x+c \\
& \text {, } \frac{1}{2}\left[u+\frac{2 \sin u \cos u}{2}\right]=x+c \\
& \frac{1}{2}[u+\sin u \cos u]=x+c \\
& \frac{1}{2}[x+y+\sin (x+y) \cos (x+y)]=x+c \quad[\because u=x+y] \\
&
\end{aligned}
$