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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{e ^{t} t}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{e ^{t} t}{2}

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{e ^{t} t}{2}

(d^{2} + 5 d + 6) y = (d + 1) 6 t^{2}

y^{^{''}} + y = 8 e^{- 6 x}

(d^{2} - 1) \cdot y = x \cdot e^{x} + cos^{2} (x) + x^{2}

y^{^{''}} + 2 y^{^{'}} + y = 8 e^{x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = x^{2} + 1

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