Differential Equation Exercise 11

Let’s look at the following differential equation:

$$ 3y^2 \cdot y' + x = 0 $$

This is a first-order differential equation because the highest derivative that appears is the first derivative, $ y' $.

It’s not linear, since the unknown function appears squared, $ y^2 $.

We can solve it by rewriting it as a separable differential equation.

First, divide both sides by $ 3y^2 $ to isolate $ y' $:

$$ \frac{3y^2 \cdot y'}{3y^2} + \frac{x}{3y^2} = \frac{0}{3y^2} $$

$$ y' + \frac{x}{3y^2} = 0 $$

At this point, the equation has the form y' + a(x)·g(y) = 0, where $ a(x) = x/3 $ and $ g(y) = 1/y^2 $.

This makes it straightforward to apply the separation of variables method.

First, rewrite the derivative in the form dy/dx:

$$ \frac{dy}{dx} + \frac{x}{3y^2} = 0 $$

Now separate the variables, placing all terms in $ y $ on one side and all terms in $ x $ on the other:

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

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

Integrate both sides with respect to their respective variables:

$$ \int y^2 \, dy = \int - \frac{x}{3} \, dx $$

$$ \int y^2 \, dy = - \int \frac{x}{3} \, dx $$

The integral on the right evaluates to $ x^2/6 + c $:

$$ \int y^2 \, dy = - \left( \frac{x^2}{6} + c \right) $$

$$ \int y^2 \, dy = - \frac{x^2}{6} - c $$

Since $ c $ is an arbitrary constant, writing $-c$ instead of $+c$ makes no difference.

For convenience, we’ll write it as $ +c $:

$$ \int y^2 \, dy = - \frac{x^2}{6} + c $$

The integral on the left is $ y^3/3 $:

$$ \frac{y^3}{3} = - \frac{x^2}{6} + c $$

As we already have a constant $ c $ in the equation, there’s no need to introduce another.

Multiply both sides by 3 to simplify:

$$ \frac{y^3}{3} \cdot 3 = \left( - \frac{x^2}{6} + c \right) \cdot 3 $$

$$ y^3 = - \frac{x^2}{6} \cdot 3 + c \cdot 3 $$

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

Since $ c $ is arbitrary, replacing $ 3c $ with $ c $ is perfectly fine.

Thus:

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

Finally, take the cube root of both sides to solve for $ y $:

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

This gives the general solution:

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

And that completes the solution.