Question

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Answer 1

(a)

Let, $\displaystyle\mathrm{y=vx}$ such that

$\displaystyle\quad \mathrm{\frac{dy}{dx}=v+x\frac{dv}{dx}}$

Then, $\displaystyle\mathrm{\frac{dy}{dx}=\frac{y}{x}+tan\frac{y}{x}}$

$\displaystyle\Rightarrow \mathrm{v+x\frac{dv}{dx}=v+tanv}$

$\displaystyle\Rightarrow \mathrm{x\frac{dv}{dx}=tanv}$

$\displaystyle\Rightarrow \mathrm{cotv\:dv=\frac{dx}{x}}$

On integration, we have

$\displaystyle\quad \mathrm{\int cotv\:dv=\int \frac{dx}{x}}$

$\displaystyle\Rightarrow \mathrm{log(sinv)=logx+logc}$

$\displaystyle\quad\quad$where c = constant of integration

$\displaystyle\Rightarrow \mathrm{sinv=cx}$

$\displaystyle\Rightarrow \mathrm{sin\frac{y}{x}=cx}$

When $\displaystyle\mathrm{x=1,\quad y=\frac{\pi}{2}}$

$\displaystyle\implies \mathrm{sin\frac{\pi}{2}=c}$

$\displaystyle\Rightarrow \mathrm{c=1}$

We have: $\displaystyle\mathrm{sin\frac{y}{x}=x}$

For $\displaystyle\mathrm{x=\frac{1}{2}}$, we have

$\displaystyle\quad \mathrm{sin2y=\frac{1}{2}}$

$\displaystyle\Rightarrow \mathrm{2y=sin^{-1}\frac{1}{2}}$

$\displaystyle\Rightarrow \mathrm{2y=\frac{\pi}{6}}$

$\displaystyle\Rightarrow \boxed{\quad\mathrm{y=\frac{\pi}{12}}\quad}$

This is the required value of y under the given conditions.

(b)

Now, $\displaystyle\mathrm{(1+x)\frac{dy}{dx}-xy=1}$

$\displaystyle\Rightarrow \mathrm{\frac{dy}{dx}-\frac{x}{x+1}y=\frac{1}{x+1}\quad .....(1)}$

$\displaystyle\therefore \mathrm{I.F.=e^{-\int \frac{x}{x+1}dx}}$

$\displaystyle\quad\mathrm{=e^{-\int \frac{(x+1)-1}{x+1}dx}}$

$\displaystyle\quad\mathrm{=e^{-\int dx+\int \frac{dx}{x+1}}}$

$\displaystyle\quad\mathrm{=e^{-x+log(x+1)}}$

$\displaystyle\quad\mathrm{=e^{-x}e^{log(x+1)}}$

$\displaystyle\quad\mathrm{=e^{-x}(x+1)}$

Multiplying (1) by I.F., we get

$\displaystyle\quad\mathrm{\frac{d}{dx}\{e^{-x}(x+1)y\}=e^{-x}}$

$\displaystyle\Rightarrow \mathrm{d\{e^{-x}(x+1)y\}=e^{-x}dx}$

On integration, we have

$\displaystyle\quad\mathrm{\int d\{e^{-x}(x+1)y\}=\int e^{-x}dx}$

$\displaystyle\Rightarrow \mathrm{e^{-x}(x+1)y=-e^{-x}+c}$

$\displaystyle\quad\quad$where c = constant of integration

When $\displaystyle\mathrm{x=0,\quad y=-1}$

$\displaystyle\implies \mathrm{-1=-1+c}$

$\displaystyle\Rightarrow \mathrm{c=0}$

We have: $\displaystyle\mathrm{e^{-x}(x+1)y=-e^{-x}}$

$\displaystyle\Rightarrow \mathrm{(x+1)y=-1\quad [\because e^{-x}\neq 0]}$

For $\displaystyle\mathrm{x=\frac{1}{2}}$, we have

$\displaystyle\quad \mathrm{\frac{3}{2}y=-1}$

$\displaystyle\Rightarrow \boxed{\quad\mathrm{y=-\frac{2}{3}}\quad}$

This is the required value of y under the given conditions.