Example Solve the following initial value problem．

$$xy^{\prime} + 2y = x^{2}, y(1) = 1$$

Answer This differential equation is 1st-order linear．Write this in the standard form.

$$y^{\prime} + \frac{2}{x}y = x$$

Integrating factor $\mu$ is

$$\mu = e^{\int \frac{2}{x}}dx = e^{2\log{x}} = x^{2} .$$

Multiply $\mu$ to both sides of the equation.

$$x^{2}y^{\prime} + 2xy = x^{3} .$$

Now the left hand side can be expressed by the derivative of the product of the integrating factor and a dependent variable $y$.

$$\frac{d(x^{2}y)}{dx} = x^{3} .$$

Now integrate with respect to $x$ to get

$$x^{2}y = \int x^{3}dx = \frac{x^{4}}{4} + c .$$

Thus, the general solution is as follows:

$$y = \frac{x^{2}}{4} + \frac{c}{x^{2}}$$

We now apply the initial condition $y(1) = 1$ to obtain

$$y = \frac{x^{2}}{4} - \frac{3}{4x^{2}}$$