What is the solution for y^{''}-y^'=5e^{2x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'=5e^{2x}

y^{^{''}} - y^{^{'}} = 5 e^{2 x}

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

Solution steps

y^{''} (x) - y^{'} (x) = 5 e^{2 x}

Solve linear ODE:

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

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

y^{^{''}} - y - 2 y = 1 + t

y^{^{''}} + 5 y = g (t), y (0) = - 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 sin (2 t), y (0) = 0, y^{^{'}} (0) = 1

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

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