What is the solution for v^{''}-1v=2e^{4x} ?

The solution for v^{''}-1v=2e^{4x} is v(x)=c_{1}e^x+c_{2}e^{-x}+2/15 e^{4x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

v^{''}-1v=2e^{4x}

v^{^{''}} - 1 v = 2 e^{4 x}

v (x) = c_{1} e^{x} + c_{2} e^{- x} + \frac{2}{15} e^{4 x}

Solution steps

v^{''} (x) - 1 v (x) = 2 e^{4 x}

Solve linear ODE:

v (x) = c_{1} e^{x} + c_{2} e^{- x} + \frac{2}{15} e^{4 x}

v (x) = c_{1} e^{x} + c_{2} e^{- x} + \frac{2}{15} e^{4 x}

y^{^{''}} + y^{^{'}} - 2 y = u π (t) - u^{2} π (t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 5

y^{^{''}} - 4 y^{^{'}} - 5 y = 6 x^{2} + 4 x + 1

y^{^{''}} + 2 y^{^{'}} + 2 y = 4

y^{^{''}} - 8 y^{^{'}} - 4 y - 2 = 2 x e^{x} sin (x)

What is the solution for v^{''}-1v=2e^{4x} ?
The solution for v^{''}-1v=2e^{4x} is v(x)=c_{1}e^x+c_{2}e^{-x}+2/15 e^{4x}