Question

What is a solution to the differential equation `y'+y/x^2=2/x^3`?

Answer 1

`y = 2 (1/x + 1) +C e^{ 1/x}`

Explanation:

`y'+y/x^2=2/x^3`

let's try separating,...., `y'=2/x^3 - y/x^2` ie NOT separable, so in wrong category

BUT we can still do it! by writing it in this form we look for the Integrating Factor `eta(x)`

`y'+ color{blue}{1/x^2} y =2/x^3`

So here `eta(x) = e^{ int color{blue}{ 1/x^2 }\ dx} = e^{ - 1/x}`

`e^{ - 1/x} y'+ e^{ - 1/x} color{blue}{1/x^2} y =e^{ - 1/x}2/x^3`

or

`d/dx (y e^{ - 1/x} ) =e^{ - 1/x}2/x^3`

`int \ d/dx (y e^{ - 1/x} ) \ dx =int \ e^{ - 1/x}2/x^3 \ dx`

So

`y e^{ - 1/x} = color{red}{int \ e^{ - 1/x}2/x^3 dx} qquad triangle`

we'll try IBP on the red bit

`int \ e^{ - 1/x}2/x^3 dx`

`= int \ 1/x^2 e^{ - 1/x} * 2/x \ dx`

`= int \ d/dx( e^{ - 1/x}) * 2/x \ dx`

which by IBP: `int uv' = uv - int u' v`

`= e^{ - 1/x}2/x - int \ e^{ - 1/x} d/dx(2/x) \ dx`

`= e^{ - 1/x}2/x +2 int \ e^{ - 1/x} (1/x^2) \ dx`

`= e^{ - 1/x}2/x + 2 e^{ - 1/x} +C`

going back to `triangle`
`implies y e^{ - 1/x} = 2 e^{ - 1/x} (1/x + 1) +C`

`implies y = 2 (1/x + 1) +C e^{ 1/x}`

Answer 2

`y(x)=C_3e^{1/x}+2 (1 + 1/x)`

Explanation:

This is a linear homogeneus differential equation. Its solution can be assembled as

`y(x) = y_h(x) +y_p(x)`

where

`y'_h(x)+(y_h(x))/x^2=0` and
`y'_p(x)+(y_p(x))/x^2=2/x^3`

The homogeneus solution is straightforward. Grouping

`(y'_h(x))/(y_h(x))=-1/x^2->log_e y_h(x) = C_0+1/x->y_h(x)=C_1e^{1/x}`

For the obtention of `y_p(x)` we will use the Lagrange's variation of constants technique.

Supposing that `y_p(x) = C(x)e^{1/x}` and substituting in the complete equation we obtain

`C'(x) = (2e^{-1/x})/x^3->C(x)=2 e^(-1/x) (1 + 1/x) + C_2`

so

`y_p(x) = 2 (1 + 1/x) + C_2e^{1/x}`

Finally

`y(x) = (C_1+C_2)e^{1/x}+2 (1 + 1/x) = C_3e^{1/x}+2 (1 + 1/x)`