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

The solution for y^{''}-3y^'+2y=e^{3t},y(0)=-1,y^'(0)=-4 is y=(-5/2-3/2)e^{2t}+5/2 e^t+1/2 e^{3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'+2y=e^{3t},y(0)=-1,y^'(0)=-4

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 t}, y (0) = - 1, y^{^{'}} (0) = - 4

y = (- \frac{5}{2} - \frac{3}{2}) e^{2 t} + \frac{5}{2} e^{t} + \frac{1}{2} e^{3 t}

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = e^{3 t}

Solve linear ODE:

y = (- \frac{5}{2} - \frac{3}{2}) e^{2 t} + \frac{5}{2} e^{t} + \frac{1}{2} e^{3 t}

y = (- \frac{5}{2} - \frac{3}{2}) e^{2 t} + \frac{5}{2} e^{t} + \frac{1}{2} e^{3 t}

y^{^{''}} - 2 y^{^{'}} + y = x^{- \frac{1}{3}} e^{x}

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

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

y^{^{''}} + 16 y = 7 x, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 cos (x)

What is the solution for y^{''}-3y^'+2y=e^{3t},y(0)=-1,y^'(0)=-4 ?
The solution for y^{''}-3y^'+2y=e^{3t},y(0)=-1,y^'(0)=-4 is y=(-5/2-3/2)e^{2t}+5/2 e^t+1/2 e^{3t}