What is the solution for y^{''}-4y^'+4y=(24x^2+2)e^{2x} ?

The solution for y^{''}-4y^'+4y=(24x^2+2)e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+2e^{2x}x^4+e^{2x}x^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-4y^'+4y=(24x^2+2)e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = (24 x^{2} + 2) e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 e^{2 x} x^{4} + e^{2 x} x^{2}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = (24 x^{2} + 2) e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 e^{2 x} x^{4} + e^{2 x} x^{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 e^{2 x} x^{4} + e^{2 x} x^{2}

What is the solution for y^{''}-4y^'+4y=(24x^2+2)e^{2x} ?
The solution for y^{''}-4y^'+4y=(24x^2+2)e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+2e^{2x}x^4+e^{2x}x^2