What is the solution for y^{''}-4y=8e^{2x} ?

The solution for y^{''}-4y=8e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+2xe^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y=8e^{2x}

y^{^{''}} - 4 y = 8 e^{2 x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + 2 x e^{2 x}

Solution steps

y^{''} (x) - 4 y = 8 e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + 2 x e^{2 x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + 2 x e^{2 x}

x \to 3 lim (x + 12)

\int_{0}^{2 π} sin (θ) d θ

\frac{d y}{d x} = e^{x + y + 3}

4 x^{3} - 13 x^{2} + 4

9 (sin (x) + x cos (x))

What is the solution for y^{''}-4y=8e^{2x} ?
The solution for y^{''}-4y=8e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+2xe^{2x}