Let y=y(x) be the solution of the differential equation x√1-x^2dy+(y√1-x^2-x ^-1x)dx=0,x (…

Question

Let $y=y(x)$ be the solution of the differential equation $x\sqrt{1-{x}^{2}}dy+(y\sqrt{1-{x}^{2}}-x{\cos }^{-1}x)dx=0,x\in (0,1),\underset{x	o {1}^{-}}{lim}y(x)=1$. Then $y(\frac{1}{2})$ equals :

Accepted Answer

$(x\sqrt{1-{x}^{2}})dy+(y\sqrt{1-{x}^{2}}-x{\cos }^{-1}x)dx=0$

$\Rightarrow (x\sqrt{1-{x}^{2}})\frac{dy}{dx}+(y\sqrt{1-{x}^{2}}-x{\cos }^{-1}x)=0$

$\Rightarrow$ $\frac{dy}{dx}+\frac{(y\sqrt{1-{x}^{2}}-x{\cos }^{-1}x)}{(x\sqrt{1-{x}^{2}})}=0$

$\Rightarrow$ $\frac{dy}{dx}+\frac{y}{x}=\frac{{\cos }^{-1}x}{\sqrt{1-{x}^{2}}}\ldots .$(i)

Bernoulli's linear ODE of first order :

$\frac{dy}{dx}+P(x)\cdot y=Q(x)$ having $I\cdot F={e}^{\int P(x)dx}$

$\begin{gathered} P(x) Q(x) \ \frac{1}{x} \frac{{\cos }^{-1}x}{\sqrt{1-{x}^{2}}} $

$\Rightarrow I.F={e}^{\int P(x)dx}={e}^{\int \frac{dx}{x}}={e}^{\ln x}=x\ldots ..(ii)$

Apply the I.F from equation $(ii)$ in $ODE(i)$ :

$x\frac{dy}{dx}+y=\frac{x{\cos }^{-1}x}{\sqrt{1-{x}^{2}}}\Rightarrow xdy+ydx=(\frac{x{\cos }^{-1}x}{\sqrt{1-{x}^{2}}})dx$

$\Rightarrow d(xy)=(\frac{x{\cos }^{-1}x}{\sqrt{1-{x}^{2}}})dx\Rightarrow \int d(xy)=\int (\frac{x{\cos }^{-1}x}{\sqrt{1-{x}^{2}}})dx\ldots ..(iii)$

We choose the first and second functions

$ u v \ {\cos }^{-1}x \frac{x}{\sqrt{1-{x}^{2}}} \ (iii) $

$ xy =({\cos }^{-1}x)\cdot \int (\frac{x}{\sqrt{1-{x}^{2}}})dx+\int [(\frac{1}{\sqrt{1-{x}^{2}}})\cdot \int (\frac{x}{\sqrt{1-{x}^{2}}})dx]dx \ =-({\cos }^{-1}x)\cdot \sqrt{1-{x}^{2}}-\int (\frac{1}{\sqrt{1-{x}^{2}}})\cdot (\sqrt{1-{x}^{2}})dx \ =-({\cos }^{-1}x)\cdot \sqrt{1-{x}^{2}}-\int (\frac{1}{\sqrt{1-{x}^{2}}})\cdot (\sqrt{1-{x}^{2}})dx \ =-({\cos }^{-1}x)\cdot \sqrt{1-{x}^{2}}-\int dx \ =-({\cos }^{-1}x)\cdot \sqrt{1-{x}^{2}}-x+C \end{gathered}$

$y(x)=\frac{C}{x}-1-\frac{({\cos }^{-1}x)\cdot \sqrt{1-{x}^{2}}}{x}\ldots .(iv)$ where $C$ is the arbitrary constant

$\underset{x	o {1}^{-}}{lim}y(x)=\underset{x	o {1}^{-}}{lim}[\frac{C}{x}-1-\frac{({\cos }^{-1}x)\cdot \sqrt{1-{x}^{2}}}{x}]=C-1-\frac{(0)\cdot (0)}{1}=C-1\ldots \ldots (v)$

We have the boundary condition

$\underset{x	o {1}^{-}}{lim}y(x)=1\Rightarrow C=2\ldots ..(vi)\leftarrow $ From equation $(v)$

We have solved the ODE to reach a particular solution $(C=2)$ from $(iv)y(x)=\frac{2}{x}-1-\frac{({\cos }^{-1}x)\cdot \sqrt{1-{x}^{2}}}{x}\ldots$. (vii)

Substitute $x=\frac{1}{2}$ in equation (vii) to evaluate $y(\frac{1}{2})=4-1-(2)\cdot (\frac{\pi }{3})\cdot (\frac{\sqrt{3}}{2})$

$=3-\frac{\pi }{\sqrt{3}}$

Solutions · 1