What is the solution for 25y^{''}-60y^'+36y=0 ?

The solution for 25y^{''}-60y^'+36y=0 is y=c_{1}e^{(6t)/5}+c_{2}e^{(6t)/5}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

25y^{''}-60y^'+36y=0

25 y^{^{''}} - 60 y^{^{'}} + 36 y = 0

y = c_{1} e^{\frac{6 t}{5}} + c_{2} e^{\frac{6 t}{5}} t

Solution steps

25 y^{''} (t) - 60 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{6 t}{5}} + c_{2} e^{\frac{6 t}{5}} t

y = c_{1} e^{\frac{6 t}{5}} + c_{2} e^{\frac{6 t}{5}} t

y^{^{'''}} + 7 y^{^{''}} + 19 y^{^{'}} + 13 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 4, y^{^{'}} (0) = - 9

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 13 y = 0

What is the solution for 25y^{''}-60y^'+36y=0 ?
The solution for 25y^{''}-60y^'+36y=0 is y=c_{1}e^{(6t)/5}+c_{2}e^{(6t)/5}t