What is the solution for y^{''}+4y=e^{4x} ?

The solution for y^{''}+4y=e^{4x} is y=c_{1}cos(2x)+c_{2}sin(2x)+1/20 e^{4x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=e^{4x}

y^{^{''}} + 4 y = e^{4 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{20} e^{4 x}

Solution steps

y^{''} (x) + 4 y = e^{4 x}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{20} e^{4 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{20} e^{4 x}

\int e^{e^{x}} d x

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

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

\int \frac{31 x ^{2}}{x ^{2} - 4} d x

- 1

What is the solution for y^{''}+4y=e^{4x} ?
The solution for y^{''}+4y=e^{4x} is y=c_{1}cos(2x)+c_{2}sin(2x)+1/20 e^{4x}