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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{''} (t) - y^{'} (t) = t e^{t}

Solve linear ODE:

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

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

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

y^{^{''}} - y = x sin (x)

y^{^{''}} + 4 y = - 12 sin (2 x)

y^{^{'''}} - 4 y^{^{''}} = x + x e^{4 x}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = x e^{x} - e^{- x} + 7

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