What is the solution for y^{''}-5y^'+6y=e^x ?

The solution for y^{''}-5y^'+6y=e^x is y=c_{1}e^{3x}+c_{2}e^{2x}+1/2 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-5y^'+6y=e^x

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{x}

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = e^{x}

Solve linear ODE:

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

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

\int \frac{8 x - 3}{12 x - 4 x ^{2} - 5} d x

\frac{d}{d X} (\frac{8 X ^{2} + 2 X + 8}{X})

l a pl a ce t r an s f or m \frac{t - 5}{5}

\int sin^{3} (11 x) d x

\int_{- 3}^{3} (9 - x^{2}) d x

What is the solution for y^{''}-5y^'+6y=e^x ?
The solution for y^{''}-5y^'+6y=e^x is y=c_{1}e^{3x}+c_{2}e^{2x}+1/2 e^x