What is the solution for y^{''}-3y^'+y3=0 ?

The solution for y^{''}-3y^'+y3=0 is y=e^{(3t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'+y3=0

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

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

Solution steps

y^{''} (t) - 3 y^{'} (t) + y 3 = 0

Solve linear ODE:

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

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

5 y^{^{''}} + 10 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 3 y = 0

y^{^{''}} - 5 y^{^{'}} - 50 y = 0

8 y^{^{''}} - 14 y^{^{'}} + 3 y = 0

What is the solution for y^{''}-3y^'+y3=0 ?
The solution for y^{''}-3y^'+y3=0 is y=e^{(3t)/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))