What is the solution for y^{''}-4y^'+4y=t^3 ?

The solution for y^{''}-4y^'+4y=t^3 is y=c_{1}e^{2t}+c_{2}te^{2t}+(t^3)/4+(3t^2)/4+(9t)/8+3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-4y^'+4y=t^3

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{3}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + \frac{9 t}{8} + \frac{3}{4}

What is the solution for y^{''}-4y^'+4y=t^3 ?
The solution for y^{''}-4y^'+4y=t^3 is y=c_{1}e^{2t}+c_{2}te^{2t}+(t^3)/4+(3t^2)/4+(9t)/8+3/4