Question

solve x^2y^''+5xy^'-5y=0 y1(x)=x and y^''-y=x-1 using reduction of order.

Answer 1

Assume there's a solution of the form \(y_2=vy_1\), then \({y_2}'=v'y_1+v{y_1}'\) and \({y_2}''=v''y_1'+2v'{y_1}'+v{y_1}''\). Since \(y_1=x\), you have \({y_1}'=1\) and \({y_1}''=0\), so \({y_2}'=xv'+v\) and \({y_2}''=v''+2v'\).

Plug these into the first ODE,
\[\begin{align*}x^2(v''+2v')+5x(xv'+v)-5xv&=0\\
x^2v''+7x^2v'&=0\\
v''+7v'&=0
\end{align*}\]
Set \(t=v'\), then you get the linear equation,
\[t'+7t=0\]

Answer 2

we can solve the second differential equation for y', and we get:
\[y' = x - 1 + y\]
then we can substitute that expression into your first differential equation, after that substitution, you should get this:
\[5xy' + \left( {{x^2} - 5} \right)y = {x^2} - {x^3}\]
please check my result

Answer 3

@Michele_Laino, just like before, I think you're mistaking this for system of ODEs, though neither asker has specified whether or not that's the case. I assumed these are two separate ODEs.