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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(\frac{h}{2})^{2}

\int_{0}^{4} (64 x - 8 x^{2}) d x

\int \frac{2 x - 8}{1 - x - x ^{2}} d x

\frac{\partial}{\partial y} (x ln (x y - 5))

\int_{0}^{9} (\frac{1}{3} x - 2) d x

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