What is the solution for y^{''}-7y^'+12y=4e^x ?

The solution for y^{''}-7y^'+12y=4e^x is y=c_{1}e^{4x}+c_{2}e^{3x}+2/3 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-7y^'+12y=4e^x

y^{^{''}} - 7 y^{^{'}} + 12 y = 4 e^{x}

y = c_{1} e^{4 x} + c_{2} e^{3 x} + \frac{2}{3} e^{x}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 12 y = 4 e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{3 x} + \frac{2}{3} e^{x}

y = c_{1} e^{4 x} + c_{2} e^{3 x} + \frac{2}{3} e^{x}

y^{^{''}} + 3 y = x^{2} sin (x)

4 y^{^{''}} + 36 y = csc (x)

y^{^{''}} - 2 y^{^{'}} + y = x e^{x} + 4, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 6 y = u^{2} (t), y (0) = 0

y^{^{''}} + 16 y = e^{5 t}

What is the solution for y^{''}-7y^'+12y=4e^x ?
The solution for y^{''}-7y^'+12y=4e^x is y=c_{1}e^{4x}+c_{2}e^{3x}+2/3 e^x