Question

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

Answer 1

`(y)/(x^2)=\sin x+\pi`

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:
`(1)/(x)\frac{\mathrm{d} y}{\mathrm{d} x}-(2y)/(x^2)=x\cos x`

We can write this equation as
`\frac{\mathrm{d} y}{\mathrm{d} x}-(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)=(-2)/(x)\,\,,\,\,q(x)=x^2\cos x`

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

we get solution as follows:

`(y)/(x^2)=\int (1)/(x^2)x^2\cos x\,dx\\(y)/(x^2)=\int \cos x\,dx\\\\(y)/(x^2)=\sin x+C`

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

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

`(y)/(x^2)=\sin x+C\\(\pi^2)/(\pi)=\sin\pi+C\\\pi=C`

So, we get solution as :

`(y)/(x^2)=\sin x+\pi`