What is the solution for y^{''}+9y=te^{2t} ?

The solution for y^{''}+9y=te^{2t} is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^{2t}t)/(13)-4/169 e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+9y=te^{2t}

y^{^{''}} + 9 y = t e^{2 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{2 t} t}{13} - \frac{4}{169} e^{2 t}

Solution steps

y^{''} (t) + 9 y = t e^{2 t}

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{2 t} t}{13} - \frac{4}{169} e^{2 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{2 t} t}{13} - \frac{4}{169} e^{2 t}

What is the solution for y^{''}+9y=te^{2t} ?
The solution for y^{''}+9y=te^{2t} is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^{2t}t)/(13)-4/169 e^{2t}