What is the solution for y^{''}-10y^'+25y=3e^{5t} ?

The solution for y^{''}-10y^'+25y=3e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(3e^{5t}t^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-10y^'+25y=3e^{5t}

y^{^{''}} - 10 y^{^{'}} + 25 y = 3 e^{5 t}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{3 e ^{5 t} t ^{2}}{2}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = 3 e^{5 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{3 e ^{5 t} t ^{2}}{2}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{3 e ^{5 t} t ^{2}}{2}

What is the solution for y^{''}-10y^'+25y=3e^{5t} ?
The solution for y^{''}-10y^'+25y=3e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(3e^{5t}t^2)/2