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

Question

Let $y = y(x)$ be the solution of the differential equation $x \frac{dy}{dx} - y = x^2 \cot x$, $x \in (0, \pi)$. If $y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$, then$6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right)$ is equal t...

Accepted Answer

The given equation is:

$x\frac{dy}{dx}-y={x}^{2}\cot x$

To get it into the standard form $\frac{dy}{dx}+P(x)y=Q(x)$, we divide the entire equation by $x$ :

$\frac{dy}{dx}-\frac{1}{x}y=x\cot x$

Here, $P(x)=-\frac{1}{x}$ and $Q(x)=x\cot x$.

Finding the Integrating Factor (I.F.)

The integrating factor is calculated as:

$={e}^{\int P(x)dx}={e}^{\int -\frac{1}{x}dx}={e}^{-\ln x}={e}^{\ln ({x}^{-1})}=\frac{1}{x}$

The solution to the differential equation is:

$\begin{gathered} y\cdot (I.F.)=\int Q(x)\cdot (I.F.)dx \ \frac{y}{x}=\int (x\cot x)\cdot \frac{1}{x}dx \ \frac{y}{x}=\int \cot xdx \ \frac{y}{x}=\ln (\sin x)+C \ y=x\ln (\sin x)+Cx \end{gathered}$

Applying the Initial Condition

We are given $y(\frac{\pi }{2})=\frac{\pi }{2}$. Plugging these values in:

$\frac{\pi }{2}=\frac{\pi }{2}\ln (\sin \frac{\pi }{2})+C(\frac{\pi }{2})$

So, the particular solution is $y=x\ln (\sin x)+x$.

We need to find the value of $6y(\frac{\pi }{6})-8y(\frac{\pi }{4})$.

Calculate $y(\frac{\pi }{6})$ :

$y(\frac{\pi }{6})=\frac{\pi }{6}\ln (\sin \frac{\pi }{6})+\frac{\pi }{6}=\frac{\pi }{6}\ln (\frac{1}{2})+\frac{\pi }{6}=\frac{\pi }{6}(1-\ln 2)$

Calculate $y(\frac{\pi }{4})$ :

$y(\frac{\pi }{4})=\frac{\pi }{4}\ln (\sin \frac{\pi }{4})+\frac{\pi }{4}=\frac{\pi }{4}\ln (\frac{1}{\sqrt{2}})+\frac{\pi }{4}=\frac{\pi }{4}\ln ({2}^{-1/2})+\frac{\pi }{4}=\frac{\pi }{4}(1-\frac{1}{2}\ln 2)$

Calculate $y(\frac{\pi }{4})$ :

$y(\frac{\pi }{4})=\frac{\pi }{4}\ln (\sin \frac{\pi }{4})+\frac{\pi }{4}=\frac{\pi }{4}\ln (\frac{1}{\sqrt{2}})+\frac{\pi }{4}=\frac{\pi }{4}\ln ({2}^{-1/2})+\frac{\pi }{4}=\frac{\pi }{4}(1-\frac{1}{2}\ln 2)$

Now, substitute these into the expression:

$\begin{gathered} 6y(\frac{\pi }{6})-8y(\frac{\pi }{4}) =6[\frac{\pi }{6}(1-\ln 2)]-8[\frac{\pi }{4}(1-\frac{1}{2}\ln 2)]=\pi (1-\ln 2)-2\pi (1-\frac{1}{2}\ln 2) \ = \pi -\pi \ln 2-2\pi +\pi \ln 2=-\pi \end{gathered}$

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