What is the solution for x^2y^{''}+3xy^'+y=0,y(1)=3.6,y^'(1)=0.4 ?

The solution for x^2y^{''}+3xy^'+y=0,y(1)=3.6,y^'(1)=0.4 is y=(3.6)/x+(4ln(x))/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^{''}+3xy^'+y=0,y(1)=3.6,y^'(1)=0.4

x^{2} y^{^{''}} + 3 x y^{^{'}} + y = 0, y (1) = 3.6, y^{^{'}} (1) = 0.4

y = \frac{3.6}{x} + \frac{4 ln ( x )}{x}

Solution steps

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

Solve linear ODE:

y = \frac{3.6}{x} + \frac{4 ln ( x )}{x}

y = \frac{3.6}{x} + \frac{4 ln ( x )}{x}

\int_{- 1}^{7} 3 x + 7 d x

\frac{\partial}{\partial x} ((x - 4) ln (x y))

\int - x cos (x) + x d x

\int_{0}^{l} cos (\frac{π y}{2 l}) d y

g (t) = t - \frac{4}{t ^{3}}

What is the solution for x^2y^{''}+3xy^'+y=0,y(1)=3.6,y^'(1)=0.4 ?
The solution for x^2y^{''}+3xy^'+y=0,y(1)=3.6,y^'(1)=0.4 is y=(3.6)/x+(4ln(x))/x