Question

Let $y=y\left(x\right)$ be the solution of the differential equation $xdy–ydx=\sqrt{x^2-y^2}dx,x\ge 1,withy\left(1\right)=0$. If the area bounded by the line $x=1,x=e^{\mathrm{\pi }},y=0andy=y\left(x\right)is\alpha e^{2\mathrm{\pi }}+\beta ,$ then the value of $10(\alpha +\beta )$ is equal to:

Answer 1

Step-1 : Solution of linear differential equation:

Given, $xdy–ydx=\sqrt{x^2-y^2}dx$

Solving the differential equation:

$$\begin{gathered} xdy–ydx=\sqrt{x^2-y^2}dx \\ \Rightarrow \frac{xdy-ydx}{x^2}=\frac{1}{x}\sqrt{1-\frac{y^2}{x^2}}dx \\ \Rightarrow \int \frac{d\left({\displaystyle \frac{y}{x}}\right)}{\sqrt{1-{\left({\displaystyle \frac{y}{x}}\right)}^2}}=\int \frac{dx}{x}\left[\because d\left(\frac{y}{x}\right)=\frac{xdy-ydx}{x^2}\right] \\ \Rightarrow {\sin }^{-1}\left(\frac{y}{x}\right)=\ln \left|x\right|+c \end{gathered}$$

Now, at $x=1,y=0$ we get:

$$\begin{gathered} {\sin }^{-1}\left(0\right)=\ln \left(1\right)+c \\ \Rightarrow c=0 \\ \therefore y=x\sin \left[\left(\ln x\right)\right] \end{gathered}$$

Step-2 : Finding the value of $\mathbf{10}\mathbf{}\mathbf{(}\mathit{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathit{\beta }\mathbf{)}$:

Now, given that $A={\int }_1^{e^{\mathrm{\pi }}}x\sin \left[\left(\ln x\right)\right]dx$

So,

$$\begin{gathered} x=e^t \\ \Rightarrow dx=e^{t}dt \\ \Rightarrow {\int }_0^{\mathrm{\pi }}{e}^{2t}\sin \left(t\right)dt=A \\ \Rightarrow \begin{array}{rcl}& & \left(\frac{e^{2t}}{5}(2\sin t-\cos t)\right)\end{array}\begin{array}{c}\mathrm{\pi }\\ \\ 0\end{array}=A\left[\because \int e^{\mathrm{ax}}\mathrm{sinbxdx}=\frac{e^{\mathrm{ax}}}{a^2+b^2}\left[\mathrm{asinbx}-\mathrm{bcosbx}\right]+c\right]\mathbf{} \end{gathered}$$

Comparing the two equations:

$\begin{array}{rcl}\alpha e^{2\mathrm{\pi }}+\beta & =& {\left(\frac{e^{2t}}{5}(2\sin t-\cos t)\right)}_0^{\mathrm{\pi }}\\ & =& \frac{1+e^{2\mathrm{\pi }}}{5}\end{array}$

$$\begin{gathered} \therefore \alpha =\frac{1}{5},\beta =\frac{1}{5} \\ \therefore 10(\alpha +\beta )=4 \end{gathered}$$

Hence, the value of $\mathbf{10}\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{)}$ is $4$.