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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{t}

y = c_{1} e^{t} + c_{2} t e^{t} - e^{t} t + e^{t} t ln (t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} - e^{t} t + e^{t} t ln (t)

y = c_{1} e^{t} + c_{2} t e^{t} - e^{t} t + e^{t} t ln (t)

a y^{^{''}} + b y^{^{'}} + c = 0

y^{^{''}} - y^{^{'}} = \frac{e ^{x}}{( 1 + e ^{x} )}

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

y^{^{''}} - y^{^{'}} + y = 10 sin (t)

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} + 2 x

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