What is the solution for y^{''}-5y^'+4y=8e^x ?

The solution for y^{''}-5y^'+4y=8e^x is y=c_{1}e^{4x}+c_{2}e^x-(8e^xx)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-5y^'+4y=8e^x

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{x}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{8 e ^{x} x}{3}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = 8 e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{8 e ^{x} x}{3}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{8 e ^{x} x}{3}

\int_{- 2}^{2} x d x

f (x) = - x^{2} + 16 x + 63, at x = 2

\int ln (4 x) d x

\int_{1}^{12} (0.5)^{x} d x

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

What is the solution for y^{''}-5y^'+4y=8e^x ?
The solution for y^{''}-5y^'+4y=8e^x is y=c_{1}e^{4x}+c_{2}e^x-(8e^xx)/3