What is the solution for x^2y^{''}+xy^'+4=0 ?

The solution for x^2y^{''}+xy^'+4=0 is y=c_{1}+c_{2}ln(x)-2ln^2(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^2y^{''}+xy^'+4=0

x^{2} y^{^{''}} + x y^{^{'}} + 4 = 0

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

Solution steps

x^{2} y^{''} (x) + x y^{'} (x) + 4 = 0

Solve linear ODE:

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

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

y^{^{''}} + 7 y^{^{'}} - 3 y = 0, y (0) = - 3, y^{^{'}} (0) = 7

\int \frac{2 x ^{\frac{3}{2}}}{3} d x

\frac{d y}{d x} = \frac{x ^{2} y ^{2}}{1 + y}

x \to 2 - lim (4 x - 5)

d er i v a t i v e y = 2 e^{x} - e^{x + 1} + e^{\frac{2}{3}} - 5 π^{2}

What is the solution for x^2y^{''}+xy^'+4=0 ?
The solution for x^2y^{''}+xy^'+4=0 is y=c_{1}+c_{2}ln(x)-2ln^2(x)