Differential Equation Exercise 15bis

Consider the differential equation:

$$ yy' - 3x^2 = 2 $$

We begin by isolating the derivative \( y' \):

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

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

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

This is a first-order equation of the form y' = f(x)g(y), where \( f(x) = 2 + 3x^2 \) and \( g(y) = 1/y \).

Such an equation can be solved using the separation of variables technique.

Separating \( x \) and \( y \) into opposite sides gives:

$$ y \cdot y' = 2 + 3x^2 $$

Rewriting \( y' \) as dy/dx:

$$ y \cdot \frac{dy}{dx} = 2 + 3x^2 $$

Moving \( dx \) to the right-hand side:

$$ y \, dy = (2 + 3x^2) \, dx $$

Integrating both sides:

$$ \int y \, dy = \int (2 + 3x^2) \, dx $$

The antiderivative of \( 2 + 3x^2 \) is \( 2x + x^3 + c \):

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

The antiderivative of \( y \) is \( y^2 / 2 \):

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

Multiplying through by 2:

$$ y^2 = 4x + 2x^3 + 2c $$

Renaming \( 2c \) as \( c \) for simplicity:

$$ y^2 = 2x^3 + 4x + c $$

Taking the square root of both sides yields:

$$ \sqrt{y^2} = \sqrt{2x^3 + 4x + c} $$

Thus, the general solution is:

$$ y = \pm \sqrt{2x^3 + 4x + c} $$

Note. This equation has no constant solution because the expression is undefined when \( y = 0 \).

That completes the solution.