What is the solution for y^{''}-8y^'+16y=2e^{4t}cos(5t) ?

The solution for y^{''}-8y^'+16y=2e^{4t}cos(5t) is y=c_{1}e^{4t}+c_{2}te^{4t}-(2e^{4t}cos(5t))/(25)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-8y^'+16y=2e^{4t}cos(5t)

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 e^{4 t} cos (5 t)

y = c_{1} e^{4 t} + c_{2} t e^{4 t} - \frac{2 e ^{4 t} cos ( 5 t )}{25}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = 2 e^{4 t} cos (5 t)

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} - \frac{2 e ^{4 t} cos ( 5 t )}{25}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} - \frac{2 e ^{4 t} cos ( 5 t )}{25}

What is the solution for y^{''}-8y^'+16y=2e^{4t}cos(5t) ?
The solution for y^{''}-8y^'+16y=2e^{4t}cos(5t) is y=c_{1}e^{4t}+c_{2}te^{4t}-(2e^{4t}cos(5t))/(25)