What is the solution for x^2y^{''}+xy^'-y=ln(x) ?

The solution for x^2y^{''}+xy^'-y=ln(x) is y=c_{1}x+(c_{2})/x-ln(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^{''}+xy^'-y=ln(x)

x^{2} y^{^{''}} + x y^{^{'}} - y = ln (x)

y = c_{1} x + \frac{c _{2}}{x} - ln (x)

Solution steps

x^{2} y^{''} (x) + x y^{'} (x) - y = ln (x)

Solve linear ODE:

y = c_{1} x + \frac{c _{2}}{x} - ln (x)

y = c_{1} x + \frac{c _{2}}{x} - ln (x)

\int \frac{x}{1 - 2 x ^{2}} d x

\frac{d}{d x} (\frac{1}{1 + 24 e ^{- 0.28 x}})

\int_{2}^{4} x^{2} d x

\int_{1}^{4} (3 x^{3} + x + 1) d x

\frac{d}{d x} (\frac{3}{8 x ^{\frac{5}{2}}})

What is the solution for x^2y^{''}+xy^'-y=ln(x) ?
The solution for x^2y^{''}+xy^'-y=ln(x) is y=c_{1}x+(c_{2})/x-ln(x)