Exercise 10.5-Additional Problems - Chapter 10 Ordinary Differential Equations Class 12 Samacheer Kalvi Solutions

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Additional Problems
Question 1.
Solve: $e^x \sqrt{1-y^2} d x+\frac{y}{x} d y=0$
Solution:
The given equation can be written as
$
x e^x d x=\frac{-y}{\sqrt{1-y^2}} d y
$
Integrating, we have
$
\begin{aligned}
\int x e^x d x & =-\int \frac{y}{\sqrt{1-y^2}} d y \\
\Rightarrow x e^x-\int e^x d x & =\frac{1}{2} \int \frac{d t}{\sqrt{t}} \text { where } t=1-y^2 \text { so that }-2 y d y=d t \\
\Rightarrow x e^x-e^x & =\frac{1}{2}\left(\frac{t^2}{1 / 2}\right)+c \\
\Rightarrow x e^x-e^x & =\sqrt{t}+c
\end{aligned}
$
$\Rightarrow x e^x-e^x-\sqrt{1-y^2}=c$ which is the required solution.
Question 2.
Solve $(x+y)^2 \frac{d y}{d x}=a^2$
Solution:

Put $\mathrm{x}+\mathrm{y}=\mathrm{z}$. Differentiating with respect to $\mathrm{x}$ we get
$
1+\frac{d y}{d x}=\frac{d z}{d x} \quad \text { i.e., } \frac{d y}{d x}=\frac{d z}{d x}-1
$
The given equation becomes $z^2\left(\frac{d z}{d x}-1\right)=a^2$
$
\Rightarrow \frac{d z}{d x}-1=\frac{a^2}{z^2} \text { or } \frac{z^2}{z^2+a^2} d z=d x
$
Integrating we have, $\int \frac{z^2}{z^2+a^2} d z=\int d x$
$
\begin{aligned}
\quad \int \frac{z^2+a^2-a^2}{z^2+a^2} d z & =x+c \Rightarrow \int\left(1-\frac{a^2}{z^2+a^2}\right) d z=x+c \\
\Rightarrow z-a^2 \cdot \frac{1}{a} \tan ^{-1} \frac{z}{a} & =x+c \\
\Rightarrow x+y-a \tan ^{-1}\left(\frac{x+y}{a}\right) & =x+c \quad(\because z=x+y) \\
\text { i.e., } y-a \tan ^{-1}\left(\frac{x+y}{a}\right) & =c, \text { which is the required solution. }
\end{aligned}
$

Question 3.
Find the cubic polynomial in $x$ which attains its maximum value 4 and minimum value 0 at $x=-1$ and 1 respectively.
Solution:
Let the cubic polynomial bey $=f(x)$. Since it attains a maximum at $=-1$ and a minimum at $x=1$. $\frac{d y}{d x}=0$ at $x=-1$ and 1
$
\frac{d y}{d x}=k(x+1)(x+1)=k\left(x^2-1\right)
$
Separating the variables we have $d y=k\left(x^2-1\right) d x$
$
\begin{aligned}
& \int d y=k \int\left(x^2-1\right) d x \\
& y=k\left(\frac{x^3}{3}-x\right)+c
\end{aligned}
$
When $\mathrm{x}=-1, \mathrm{y}=4$ and when $\mathrm{x}=1,7=0$
Substituting the equation (1) we have
$
2 \mathrm{k}+3 \mathrm{c}=12 ;-2 \mathrm{k}+3 \mathrm{c}=0
$
On solving we have $k=3$ and $c=2$. Substituting these values in (1) we get the required cubic polynomial $y=x^3-3 x+2$.

Question 4.
The normal lines to a given curve at each point $(\mathrm{x}, \mathrm{y})$ on the curve pass through the point $(2,0)$. The curve passes through the point $(2,3)$. Formulate the differential equation representing the problem and hence find the equation of the curve.
Solution:
Slope of the normal at any point $\mathrm{P}(\mathrm{x}, \mathrm{y})=-\frac{d x}{d y}$
Slope of the normal AP $=\frac{y-0}{x-2} \quad \therefore-\frac{d x}{d y}=\frac{y}{x-2} \Rightarrow y d y=(2-x) d x$
Integrating both sides, $\frac{y^2}{2}=2 x-\frac{x^2}{2}+c$
Since the curve passes through $(2,3)$
$
\begin{array}{r}
\frac{9}{2}=4-\frac{4}{2}+c \Rightarrow c=\frac{5}{2} ; \text { put } c=\frac{5}{2} \text { in }(1) \\
\frac{y^2}{2}=2 x-\frac{x^2}{2}+\frac{5}{2} \Rightarrow y^2=4 x-x^2+5
\end{array}
$