# Xd^2y/dx^2-4xdy/dx+4xy=5xe^(3x)+4xe^(2x)??

Jul 6, 2018

$y = {e}^{2 x} \left(5 {e}^{x} + 2 {x}^{2} + {c}_{1} x + {c}_{2}\right)$

#### Explanation:

$x \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x y = 5 x {e}^{3 x} + 4 x {e}^{2 x}$

Eliminate common $x$ term

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - 4 \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y = 5 {e}^{3 x} + 4 {e}^{2 x}$

Expressing in terms of linear differential operator $D = \frac{d}{\mathrm{dx}}$:

$\left({D}^{2} - 4 D + 4\right) y = 5 {e}^{3 x} + 4 {e}^{2 x}$

Factoring:

$\left(D - 2\right) {\underbrace{\left(D - 2\right) y}}_{= z} = 5 {e}^{3 x} + 4 {e}^{2 x} q \quad \triangle$

$z ' - 2 z = 5 {e}^{3 x} + 4 {e}^{2 x}$

Integrating factor: $\exp \left(\int \left(- 2\right) \mathrm{dx}\right) = {e}^{- 2 x}$

Distribute integrating factor across equation:

${\left({e}^{- 2 x} z\right)}^{'} = 5 {e}^{x} + 4$

Integrating:

${e}^{- 2 x} z = 5 {e}^{x} + 4 x + {c}_{1}$

$\therefore z = 5 {e}^{3 x} + 4 x {e}^{2 x} + {c}_{1} {e}^{2 x} q \quad \boldsymbol{= \left(D - 2\right) y}$

$y ' - 2 y = 5 {e}^{3 x} + 4 x {e}^{2 x} + {c}_{1} {e}^{2 x}$

Using same integrating factor again:

${\left({e}^{- 2 x} y\right)}^{'} = 5 {e}^{x} + 4 x + {c}_{1}$

Integrating:

${e}^{- 2 x} y = 5 {e}^{x} + 2 {x}^{2} + {c}_{1} x + {c}_{2}$

$y = 5 {e}^{3 x} + 2 {x}^{2} {e}^{2 x} + {c}_{1} x {e}^{2 x} + {c}_{2} {e}^{2 x}$

$= {e}^{2 x} \left(5 {e}^{x} + 2 {x}^{2} + {c}_{1} x + {c}_{2}\right)$