What is the solution for y^{''}-2ay^'+a^2y=0 ?

The solution for y^{''}-2ay^'+a^2y=0 is y=c_{1}e^{at}+c_{2}te^{at}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2ay^'+a^2y=0

y^{^{''}} - 2 a y^{^{'}} + a^{2} y = 0

y = c_{1} e^{a t} + c_{2} t e^{a t}

Solution steps

y^{''} (t) - 2 a y^{'} (t) + a^{2} y = 0

Solve linear ODE:

y = c_{1} e^{a t} + c_{2} t e^{a t}

y = c_{1} e^{a t} + c_{2} t e^{a t}

\int - sin (θ) d θ

\int 3 t^{2} - 2 t d t

\int_{- 1}^{2} 4 x (1 - x^{2}) d x

\frac{d}{d x} (2 1^{x})

y^{^{''}} - 4 y^{^{'}} + 26 y = 0

What is the solution for y^{''}-2ay^'+a^2y=0 ?
The solution for y^{''}-2ay^'+a^2y=0 is y=c_{1}e^{at}+c_{2}te^{at}