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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (- 0.4 x^{2} + 2.2 x + 4.7)

\frac{d}{d x} (e^{x} (tan (x) - x))

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

2 x^{3} + 3 x^{2} - 12 x + 2

\int (x^{2} - 5 x) \cdot e^{x} d x

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