What is the solution for y^{''}-6y^'+8y=e^{3t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-6y^'+8y=e^{3t},y(0)=0,y^'(0)=0 is y=(-1/2+1)e^{4t}+1/2 e^{2t}-e^{3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+8y=e^{3t},y(0)=0,y^'(0)=0

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

y = (- \frac{1}{2} + 1) e^{4 t} + \frac{1}{2} e^{2 t} - e^{3 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = e^{3 t}

Solve linear ODE:

y = (- \frac{1}{2} + 1) e^{4 t} + \frac{1}{2} e^{2 t} - e^{3 t}

y = (- \frac{1}{2} + 1) e^{4 t} + \frac{1}{2} e^{2 t} - e^{3 t}

y^{^{''}} - 25 y = 8 e^{5 x}

y^{^{''}} + 4 y = t^{2} + 2 e^{t}

y^{^{''}} + 9 y = 6 sin (3 x)

y^{^{''}} - y^{^{'}} - 2 y = e^{3 x}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + y = 6

What is the solution for y^{''}-6y^'+8y=e^{3t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-6y^'+8y=e^{3t},y(0)=0,y^'(0)=0 is y=(-1/2+1)e^{4t}+1/2 e^{2t}-e^{3t}