What is the solution for x^2y^{''}-5xy^'+13y=0 ?

The solution for x^2y^{''}-5xy^'+13y=0 is y=x^3(c_{1}cos(2ln(x))+c_{2}sin(2ln(x)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^{''}-5xy^'+13y=0

x^{2} y^{^{''}} - 5 x y^{^{'}} + 13 y = 0

y = x^{3} (c_{1} cos (2 ln (x)) + c_{2} sin (2 ln (x)))

Solution steps

x^{2} y^{''} (x) - 5 x y^{'} (x) + 13 y = 0

Solve linear ODE:

y = x^{3} (c_{1} cos (2 ln (x)) + c_{2} sin (2 ln (x)))

y = x^{3} (c_{1} cos (2 ln (x)) + c_{2} sin (2 ln (x)))

\frac{d}{d x} (sin (cos (x^{2} + 3)))

y = 4 x, y = x^{3} + 3 x^{2}

\int_{0}^{1} \frac{2}{7} x^{2} (x + 3) d x

\int_{0}^{1} π (5 x^{6} - 5 x)^{2} d x

\int_{0}^{10} x e^{- x} d x

What is the solution for x^2y^{''}-5xy^'+13y=0 ?
The solution for x^2y^{''}-5xy^'+13y=0 is y=x^3(c_{1}cos(2ln(x))+c_{2}sin(2ln(x)))