Exercise on Differential Equations

We want to solve the following first - order differential equation:

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

Let’s start by isolating y′. We divide both sides of the equation by xy:

$$ \frac{ xyy' }{xy} = \frac{ 2x^2+y^2 }{xy} $$

$$ y' = \frac{2x^2}{xy} + \frac{y^2}{xy} $$

$$ y' = \frac{2x}{y} + \frac{y}{x} $$

This is a homogeneous differential equation of the form \(y' = f(y/x)\).

To solve it, we introduce the auxiliary variable t = y/x:

$$ t = \frac{y}{x} $$

From this, we can express y and its derivative y′ with respect to x:

$$ y = t \cdot x $$

$$ y' = D[ t \cdot x ] = t'x + t $$

Now, substitute y = tx into the original differential equation:

$$ y' = \frac{2x}{y} + \frac{y}{x} $$

$$ y' = \frac{2x}{tx} + \frac{tx}{x} $$

$$ y' = \frac{2}{t} + t $$

Next, replace y′ with t′x + t in the equation:

$$ t'x + t = \frac{2}{t} + t $$

We can simplify by subtracting +t from both sides:

$$ t'x = \frac{2}{t} $$

Now, let’s rewrite t′ using the notation dt/dx:

$$ \frac{dt}{dx} \cdot x = \frac{2}{t} $$

At this point, the equation can be solved using the method of separation of variables.

We separate t and x into different sides of the equation:

$$ t \cdot dt = \frac{2}{x} \cdot dx $$

Now integrate both sides with respect to their variables:

$$ \int t \cdot dt = \int \frac{2}{x} \cdot dx $$

$$ \int t \cdot dt = 2 \cdot \int \frac{1}{x} \cdot dx $$

The integral on the right gives the antiderivative log(x) + c:

$$ \int t \cdot dt = 2 \cdot \log x + c $$

The integral on the left is \(\mathbf{t^2/2}\):

$$ \frac{t^2}{2} = 2 \cdot \log x + c $$

$$ t^2 = 4 \cdot \log x + c $$

Now, substitute back \(t = y/x\):

$$ \left(\frac{y}{x}\right)^2 = 4 \cdot \log x + c $$

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

$$ y^2 = x^2 \cdot ( 4 \cdot \log x + c ) $$

$$ y^2 = 4 \cdot x^2 \cdot \log x + c \cdot x^2 $$

Taking the square root of both sides:

$$ \sqrt{y^2} = \sqrt{4 \cdot x^2 \cdot \log x + c \cdot x^2} $$

$$ y = \sqrt{4 \cdot x^2 \cdot \log x + c \cdot x^2} $$

This is the general solution to the differential equation.

And that’s the result.