$xy'-3y=x-1$ ? Solve DE

Question

7241 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-06T07:26:26

$y(x)=ax^3-1/2x+1/3$

We have

$xdy/dx-3y=x-1$

Divide all terms by $x$

$dy/dx-(3y)/x=(x-1)/x$

The ODE is now in a form where it can be solved using an integrating factor

Let $mu=expint-3/x=exp(-3lnx)=expln(1/x^3)=1/x^3$

Now multiply the whole equation by $mu$

$1/x^3 dy/dx-(3y)/x^4=(x-1)/x^4$

The LHS factors as a derivative and we can integrate the RHS

$intd/dx(y/x^3)dx=intx^-3-x^-4dx$

$y/x^3=-1/2x^-2+1/3x^-3+a$

Solve for $y$

$y=y(x)=ax^3-1/2x+1/3$

xy'-3y=x-1xy'-3y=x-1? Solve DE