Let y=y(x) be a solution of the differential equation [(x x) d y+(x y x+y x-1) d x=0,0

Question

Let $y=y(x)$ be a solution of the differential equation $$(x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2}$$. If $$\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3}$$

Accepted Answer

Given, differential equation

$\begin{gathered} (x\cos x)dy+(xy\sin x+y\cos x-1)dx=0,0<x<\frac{\pi }{2} \ \Rightarrow \frac{dy}{dx}+\frac{xy\sin x+y\cos x-1}{x\cos x}=0 \ \Rightarrow \frac{dy}{dx}+(\frac{xy\sin x+y\cos x}{x\cos x})=\frac{1}{x\cos x} \ \Rightarrow \frac{dy}{dx}+(	an x+\frac{1}{x})y=\frac{1}{x\cos x} \end{gathered}$

Which is linear differential equation in the form of

$\begin{gathered} \frac{dy}{dx}+Py=Q \ 	herefore IF={e}^{\int (	an x+\frac{1}{x})dx}={e}^{(\log \sec x+\log x)}={e}^{\log (x\sec x)}=x\sec x \end{gathered}$

$	herefore $ The general solution of the given differential equation

$\begin{gathered} y\cdot IF=\int (Q	imes IF)dx+c \ \Rightarrow y(x\sec x)=\int (\frac{1}{x\cos x}x\sec x)dx+c \ \Rightarrow xy\sec x=\int {\sec }^{2}xdx+c \ \Rightarrow xy\sec x=	an x+c........(i) \end{gathered}$

Since, $\frac{\pi }{3}y(\frac{\pi }{3})=\sqrt{3}\Rightarrow y(\frac{\pi }{3})=\frac{3\sqrt{3}}{\pi }$

$\begin{gathered} 	herefore \frac{\pi }{3}(\frac{3\sqrt{3}}{\pi })\sec (\frac{\pi }{3})=	an (\frac{\pi }{3})+c \ \ \Rightarrow \sqrt{3}(2)=\sqrt{3}+c \ \ \Rightarrow c=\sqrt{3} \end{gathered}$

On putting the value of $c$ in Eq. (i), we get

$\begin{gathered} xy\sec x=	an x+\sqrt{3} \ \Rightarrow y=\frac{1}{x}(\sin x+\sqrt{3}\cos x) \ {y}^{'}=\frac{1}{x}(\cos x-\sqrt{3}\sin x)-\frac{1}{{x}^{2}}(\sin x+\sqrt{3}\cos x) \end{gathered}$

$\begin{gathered} {y}^{''}=\frac{1}{x}(-\sin x-\sqrt{3}\cos x)-\frac{1}{{x}^{2}}(\cos x-\sqrt{3}\sin x)-\frac{1}{{x}^{2}}(\cos x-\sqrt{3}\sin x) \ -\frac{2}{{x}^{3}}(\cos x-\sqrt{3}\sin x) \ 	herefore |\frac{\pi }{6}{y}^{''}(\frac{\pi }{6})+2{y}^{'}(\frac{\pi }{6})|=|-2|=2 \end{gathered}$

Solutions · 1