What is the solution for t^2y^{''}+3ty^'+y= 1/t ?

The solution for t^2y^{''}+3ty^'+y= 1/t is y=(c_{1})/t+(c_{2}ln(t))/t+(ln^2(t))/(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^2y^{''}+3ty^'+y= 1/t

t^{2} y^{^{''}} + 3 t y^{^{'}} + y = \frac{1}{t}

y = \frac{c _{1}}{t} + \frac{c _{2} ln ( t )}{t} + \frac{ln ^{2} ( t )}{2 t}

Solution steps

t^{2} y^{''} (t) + 3 t y^{'} (t) + y = \frac{1}{t}

Solve linear ODE:

y = \frac{c _{1}}{t} + \frac{c _{2} ln ( t )}{t} + \frac{ln ^{2} ( t )}{2 t}

y = \frac{c _{1}}{t} + \frac{c _{2} ln ( t )}{t} + \frac{ln ^{2} ( t )}{2 t}

\int_{0}^{1} x - x^{2} d x

x \to 0 lim (\frac{1}{3 + x} - \frac{1}{3})

\int 6 + 7 t^{2} d t

x \to 1 + lim (\frac{( x )}{ln ( x )})

\frac{d y}{d x} = 7 x e^{4 y}

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