What is the solution for x^2y^{''}-3xy^'+4y=0 ?

The solution for x^2y^{''}-3xy^'+4y=0 is y=c_{1}x^2+c_{2}ln(x)x^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^{''}-3xy^'+4y=0

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = 0

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

Solution steps

x^{2} y^{''} (x) - 3 x y^{'} (x) + 4 y = 0

Solve linear ODE:

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

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

ln (x), x = 4

y^{^{''}} + y = sin (x), y (0) = 1, y (\frac{π}{2}) = 0

x^{7} (1 - \frac{2}{x + 4})

\frac{d}{d x} ((2 x - 6)^{4} (x^{2} + x + 1)^{5})

\int (t + 1)^{2} d t

What is the solution for x^2y^{''}-3xy^'+4y=0 ?
The solution for x^2y^{''}-3xy^'+4y=0 is y=c_{1}x^2+c_{2}ln(x)x^2