What is the solution for x^3y^'+3x^2y= 1/x ,y(1)=3 ?

The solution for x^3y^'+3x^2y= 1/x ,y(1)=3 is y=(ln(x))/(x^3)+3/(x^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^3y^'+3x^2y= 1/x ,y(1)=3

x^{3} y^{^{'}} + 3 x^{2} y = \frac{1}{x}, y (1) = 3

y = \frac{ln ( x )}{x ^{3}} + \frac{3}{x ^{3}}

Solution steps

x^{3} y^{'} (x) + 3 x^{2} y = \frac{1}{x}

Solve linear ODE:

y = \frac{ln ( x )}{x ^{3}} + \frac{3}{x ^{3}}

y = \frac{ln ( x )}{x ^{3}} + \frac{3}{x ^{3}}

\frac{\partial}{\partial x} (x sin (y) sin (z))

\int (x^{2} + 1)^{\frac{1}{2}} x d x

\int (- 3 x + 4) e - x d x

\int_{0}^{\frac{π}{2}} 7 cos^{2} (x) d x

y = x^{2}, y = - x^{2} + 18 x

What is the solution for x^3y^'+3x^2y= 1/x ,y(1)=3 ?
The solution for x^3y^'+3x^2y= 1/x ,y(1)=3 is y=(ln(x))/(x^3)+3/(x^3)