Differential Equation Exercise 14

We are given the differential equation:

$$ 3y^2y' = x+1 $$

First, we isolate the derivative $ y' $:

$$ \frac{3y^2y'}{3y^2} = \frac{x+1}{3y^2} $$

$$ y' = \frac{x+1}{3y^2} $$

This is now in the form y' = f(x)g(y), where $ f(x) = \frac{x+1}{3} $ and $ g(y) = \frac{1}{y^2} $:

$$ y' = \frac{x+1}{3} \cdot \frac{1}{y^2} $$

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

$$ y' = \frac{x+1}{3y^2} $$

Rewriting $ y' $ as dy/dx and moving dx to the other side:

$$ \frac{dy}{dx} = \frac{x+1}{3y^2} $$

$$ dy = \frac{x+1}{3y^2} \, dx $$

Separating the variables $ x $ and $ y $ into opposite sides of the equation:

$$ 3y^2 \, dy = (x+1) \, dx $$

Integrating both sides:

$$ \int 3y^2 \, dy = \int (x+1) \, dx $$

$$ 3 \cdot \int y^2 \, dy = \int (x+1) \, dx $$

The antiderivative of $ x+1 $ is $ \frac{x^2}{2} + x + c $:

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

The antiderivative of $ y^2 $ is $ \frac{y^3}{3} $:

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

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

Solving for $ y $ by taking the cube root:

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

We thus obtain the general solution to the differential equation:

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

Next, we check for any possible constant solutions by setting $ y = 0 $:

$$ y' = \frac{x+1}{3y^2} $$

Since the expression is undefined for $ y = 0 $, no constant solution exists in this case.

Therefore, the solution is:

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

And that completes the solution.