Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Answer:[tex]\frac{y}{x^2}=\sin x+\pi[/tex]Step-by-step explanation:Consider linear differential equation [tex]\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)[/tex]It's solution is of form [tex]y\,I.F=\int I.F\,q(x)\,dx[/tex] where I.F is integrating factor given by [tex]I.F=e^{\int p(x)\,dx}[/tex].Given: [tex]\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x[/tex]We can write this equation as [tex]\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x[/tex]On comparing this equation with [tex]\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)[/tex], we get [tex]p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x[/tex]I.F = [tex]e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}[/tex]      { formula used: [tex]\ln a^b=b\ln a[/tex] }we get solution as follows:[tex]\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C[/tex]{ formula used: [tex]\int \cos x\,dx=\sin x[/tex] }Applying condition:[tex]y(\pi)=\pi^2[/tex][tex]\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C[/tex]So, we get solution as :[tex]\frac{y}{x^2}=\sin x+\pi[/tex]