What is the solution for y^{''}-6y^'+8y=2,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-6y^'+8y=2,y(0)=0,y^'(0)=0 is y=(1/2-1/4)e^{4t}-1/2 e^{2t}+1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+8y=2,y(0)=0,y^'(0)=0

y^{^{''}} - 6 y^{^{'}} + 8 y = 2, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{1}{2} - \frac{1}{4}) e^{4 t} - \frac{1}{2} e^{2 t} + \frac{1}{4}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = 2

Solve linear ODE:

y = (\frac{1}{2} - \frac{1}{4}) e^{4 t} - \frac{1}{2} e^{2 t} + \frac{1}{4}

y = (\frac{1}{2} - \frac{1}{4}) e^{4 t} - \frac{1}{2} e^{2 t} + \frac{1}{4}

x^{^{''}} + 49 x = 52 cos (6 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = - 2 x

y^{^{''}} - 5 y^{^{'}} + 6 y = 12 x - 4

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e^{3 x}

What is the solution for y^{''}-6y^'+8y=2,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-6y^'+8y=2,y(0)=0,y^'(0)=0 is y=(1/2-1/4)e^{4t}-1/2 e^{2t}+1/4