What is the solution for y^{''}-3y^'+18y=0 ?

The solution for y^{''}-3y^'+18y=0 is y=e^{(3t)/2}(c_{1}cos((3sqrt(7)t)/2)+c_{2}sin((3sqrt(7)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'+18y=0

y^{^{''}} - 3 y^{^{'}} + 18 y = 0

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{3 7 t}{2}) + c_{2} sin (\frac{3 7 t}{2}))

Solution steps

y^{''} (t) - 3 y^{'} (t) + 18 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{3 7 t}{2}) + c_{2} sin (\frac{3 7 t}{2}))

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{3 7 t}{2}) + c_{2} sin (\frac{3 7 t}{2}))

\int (x - 2) (x - 3)^{6} d x

\frac{\partial}{\partial x} (4 (x - y) e^{4 y + 2 x^{2}})

f (x) = x^{3} - 1

\frac{\partial}{\partial x} (4 x^{3} - 5 x^{2} + 8 x)

y^{^{'}} = \frac{2 x + y}{2 x + y + 3}

What is the solution for y^{''}-3y^'+18y=0 ?
The solution for y^{''}-3y^'+18y=0 is y=e^{(3t)/2}(c_{1}cos((3sqrt(7)t)/2)+c_{2}sin((3sqrt(7)t)/2))