# a second order ordinary differential equation to be solved by method of variation of parameters

how to solve $$x^2y"-2x(1-x)y'+2(x+1)y=x^3$$ by method of variation of parameters. how do you find solution of the above equation? iam familiar with method of variation of parameters but dont know how to find $$y_1$$ and $$y_2.$$ please help.

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Jan 3, 2021 at 10:23
• For to find $y_{1}$ and $y_{2}$ you need to solve the homogeneous-equation: $x^{2}y''-2x(1-x)y'+2(x+1)y=0$ for to solve this I think you can read here: mathworld.wolfram.com/Sturm-LiouvilleEquation.html since that we can re-write the homogeneous equation a: $\frac{d}{dx}\left(\frac{e^{2x}y'}{x^{2}}\right)+\frac{2e^{2x}(1+x)y}{x^{4}}=0$.
– user798113
Jan 3, 2021 at 10:32
• Note that a solution for the homogeneous equation is: $y_{1}=e^{-2x}(x-1)x^{2}$ and you can find the other solution using the Abel's formula: fac-staff.seattleu.edu/oliveras/web/teaching/DiffEq/Notes/…
– user798113
Jan 3, 2021 at 10:39
• Are you familiar with how to deal with equations with variable coefficients? Do you know the method of Frobenius? Jan 3, 2021 at 10:43
• anyone please solve this. Jan 3, 2021 at 12:10

You might first need to set an interval for $$x$$, as the equation degenerates as $$x\rightarrow 0$$.