What is the solution for y^{''}+10y^'+25y=7e^{2x} ?

The solution for y^{''}+10y^'+25y=7e^{2x} is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+1/7 e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+10y^'+25y=7e^{2x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 7 e^{2 x}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{1}{7} e^{2 x}

Solution steps

y^{''} (x) + 10 y^{'} (x) + 25 y = 7 e^{2 x}

Solve linear ODE:

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{1}{7} e^{2 x}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{1}{7} e^{2 x}

\int_{- \infty}^{3} x e^{x} d x

5 e^{x} cos (x) (0.5)

\int x sin (7 x) d x

(1 + cos^{3} (x^{7}))^{12}

\frac{\partial}{\partial x} (y^{3} sin (6 x))

What is the solution for y^{''}+10y^'+25y=7e^{2x} ?
The solution for y^{''}+10y^'+25y=7e^{2x} is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+1/7 e^{2x}