What is the solution for 4y^{''}-20y^'+25y=0 ?

The solution for 4y^{''}-20y^'+25y=0 is y=c_{1}e^{(5t)/2}+c_{2}e^{(5t)/2}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}-20y^'+25y=0

4 y^{^{''}} - 20 y^{^{'}} + 25 y = 0

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

Solution steps

4 y^{''} (t) - 20 y^{'} (t) + 25 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (x^{2} 4 - x)

(- 3.1) (2.1)

x^{^{'}} = 6 - \frac{3}{5} \cdot x

\int e^{x} tan (e^{x}) d x

\frac{d}{d x} (a sin (2 x) + b cos (2 x))

What is the solution for 4y^{''}-20y^'+25y=0 ?
The solution for 4y^{''}-20y^'+25y=0 is y=c_{1}e^{(5t)/2}+c_{2}e^{(5t)/2}t