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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = 2 e^{2 t} - 2 e^{t} - 2 e^{t} t

Solution steps

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

Solve linear ODE:

y = 2 e^{2 t} - 2 e^{t} - 2 e^{t} t

y = 2 e^{2 t} - 2 e^{t} - 2 e^{t} t

x \to 0 lim (\frac{1}{x} - 1)

\int e^{l n (x^{2})} d x

\int sec^{2} (5 x) tan^{6} (5 x) d x

f (x) = x^{2} - 49 x - 456, at x = 2

\frac{d}{d x} (1 + cot^{2} (x))

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