What is the solution for y^{''}-4y^'+4y=5t^2 ?

The solution for y^{''}-4y^'+4y=5t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+(5t^2)/4+(5t)/2+15/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-4y^'+4y=5t^2

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{2}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 5 t^{2}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

What is the solution for y^{''}-4y^'+4y=5t^2 ?
The solution for y^{''}-4y^'+4y=5t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+(5t^2)/4+(5t)/2+15/8