Differential Equation Exercise 10

In this exercise, we solve the first-order differential equation:

$$ 2yy' - x^2 = 0 $$

Dividing through by \( 2y \) to isolate \( y' \):

$$ \frac{2yy'}{2y} - \frac{x^2}{2y} = 0 $$

$$ y' - \frac{x^2}{2y} = 0 $$

This is a separable differential equation of the form \( y' + f(x)g(y) = 0 \), where \( f(x) = -x^2 \) and \( g(y) = 2y \).

Expressing \( y' \) in derivative notation:

$$ \frac{dy}{dx} - \frac{x^2}{2y} = 0 $$

Separating the variables, with \( y \) on the left and \( x \) on the right:

$$ \frac{dy}{dx} = \frac{x^2}{2y} $$

$$ 2y \, dy = x^2 \, dx $$

Integrating both sides:

$$ \int 2y \, dy = \int x^2 \, dx $$

The integral of \( x^2 \) is \( \frac{x^3}{3} + c \):

$$ \int 2y \, dy = \frac{x^3}{3} + c $$

The integral of \( 2y \) is \( y^2 \):

$$ y^2 = \frac{x^3}{3} + c $$

Taking the square root of both sides yields:

$$ \sqrt{y^2} = \sqrt{\frac{x^3}{3} + c} $$

Thus, the general solution is:

$$ y = \sqrt{\frac{x^3}{3} + c} $$