Answer to Question #293047 in Differential Equations for jacob

Question

 3. Find the particular solution of the linear equation dy/dx + 1/x y = x if y(1) = 0

Answer 1

$$y'+\dfrac{1}{x}y=x$$

Integration factor

$$\mu(x)=e^{\int(dx/x)}=x$$

$$xy'+y=x^2$$

$$d(xy)=x^2 dx$$

Integrate

$$\int d(xy)=\int x^2 dx$$

$$xy=\dfrac{x^3}{3}+C$$

$$y=\dfrac{x^2}{3}+\dfrac{C}{x}$$

Given $y(1)=0$

$$0=\dfrac{(1)^2}{3}+\dfrac{C}{1}=>C=-\dfrac{1}{3}$$

The solution of the initial value problem is

$$y=\dfrac{x^2}{3}-\dfrac{1}{3x}$$