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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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