# Show that y_1=x^2 is a solution to the differential equation x^2y'' -(x^2+4x)y' +(2x+6)y=0 ?

May 10, 2018

We seek to show that ${y}_{1} = {x}^{2}$ satisfies:

${x}^{2} y ' ' - \left({x}^{2} + 4 x\right) y ' + \left(2 x + 6\right) y = 0$

Differentiating the given function we have:

${y}_{1} ' \setminus \setminus = 2 x$
${y}_{1} ' ' = 2$

And substituting into the given ODE, we have:

${x}^{2} y ' ' - \left({x}^{2} + 4 x\right) y ' + \left(2 x + 6\right) y$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {x}^{2} \left(2\right) - \left({x}^{2} + 4 x\right) \left(2 x\right) + \left(2 x + 6\right) \left({x}^{2}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 {x}^{2} - 2 {x}^{3} - 8 {x}^{2} + 2 {x}^{3} + 6 {x}^{2}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0 \setminus \setminus \setminus \setminus$ QED