Question

Find the general equation of the following differential equations:

xy' + (1 + 2x^2)y = x3e-x^2

xy' +2y = (2/x^2) + 1

y' + 4y/(x-1) = 1/(x-1)5 + sinx/(x-1)4

Answer 1

Each of these ODEs is linear. Isolate
`y'` and find the corresponding integrating factors.

• 
  `xy'+(1+2x^2)y=x^3e^(-x^2)`

`\implies y'+\frac{1+2x^2}xy=x^2e^(-x^2)`

The integrating factor in this case is
`\mu` where

`\ln\mu=\displaystyle\int\frac{1+2x^2}x\,\mathrm dx=\int\left(\frac1x+2x\right)\,\mathrm dx`

`\implies\ln\mu=\ln x+x^2\implies\mu=xe^(x^2)`

Multiplying both sides of the ODE by
`\mu` gives

`xe^(x^2)y'+(1+2x^2)e^(x^2)y=x^3`

Notice that
`(xe^(x^2))'=(1+2x^2)e^(x^2)`; condensing the left side gives

`(xe^(x^2)y)'=x^3`

`\implies xe^(x^2)y=\frac{x^4}4+C`

`\implies\boxed{y=(x^3)/(4e^(x^2))+Ce^(-x^2)}`

• 
  `xy'+2y=2x^(-2)+1`

`\implies y'+\frac2xy=\frac2{x^2}+1`

`\ln\mu=\displaystyle\int\frac2x\,\mathrm dx=2\ln x\implies\mu=x^2`

`x^2y'+2xy=2+x^2\implies(x^2y)'=2+x^2\implies x^2y=2x+\frac{x^3}3+C`

`\implies\boxed{y=\frac2x+\frac x3+\frac C{x^2}}`

• 
  `y'+4(x-1)^(-1)y=(x-1)^(-5)+(x-1)^(-4)\sin x`

`\ln\mu=\displaystyle\int\frac4{x-1}\,\mathrm dx=4\ln(x-1)\implies\mu=(x-1)^4`

`(x-1)^4y'+4(x-1)^3y=\frac1{x-1}+\sin x`

`\implies((x-1)^4y)'=\frac1{x-1}+\sin x`

`\implies(x-1)^4y=\ln|x-1|-\cos x+C`

`\implies\boxed-\cos x+C)/((x-1)^4)`