What is the solution for y^{''}-6y^'+8y=e^{4x} ?

The solution for y^{''}-6y^'+8y=e^{4x} is y=c_{1}e^{4x}+c_{2}e^{2x}+(e^{4x}x)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+8y=e^{4x}

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{4 x}

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = e^{4 x}

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} = 3 sin (3 t)

y^{^{''}} + y^{^{'}} = - cos (2 x)

y^{^{''}} - 3 y^{^{'}} + 5 = e^{2 x} - 3 cos (3 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 x

x^{^{''}} + 9 x = sin (t), x (0) = 1, x^{^{'}} (0) = 1

What is the solution for y^{''}-6y^'+8y=e^{4x} ?
The solution for y^{''}-6y^'+8y=e^{4x} is y=c_{1}e^{4x}+c_{2}e^{2x}+(e^{4x}x)/2