Question

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

Answer 1

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$.

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$, we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$      { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition:$y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$