We are tasked with solving the first-order differential equation:

$y' + y = x y^{2/3}$

Step 1: Simplifying the Equation

First, we rewrite the equation for clarity:

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

This is not linear due to the $y^{2/3}$ term, but we will attempt a substitution to simplify the equation.

Step 2: Substitution

Let us use the substitution:

$y = u^3$

Then, $y' = 3u^2 \frac{du}{dx}$. Substituting into the original equation gives:

$3u^2 \frac{du}{dx} + u^3 = x u^2$

Step 3: Simplify the New Equation

We can divide the whole equation by $u^2$ (assuming $u \neq 0$):

$3 \frac{du}{dx} + u = x$

This is now a linear first-order differential equation.

Step 4: Solve the Linear Differential Equation

The equation $3 \frac{du}{dx} + u = x$ can be solved using an integrating factor.

The integrating factor is:

$\mu(x) = e^{\int \frac{1}{3} dx} = e^{x/3}$

Multiply both sides of the equation by $e^{x/3}$:

$e^{x/3} \cdot 3 \frac{du}{dx} + e^{x/3} \cdot u = e^{x/3} \cdot x$

The left side becomes the derivative of $3u e^{x/3}$, so:

$\frac{d}{dx} \left( 3u e^{x/3} \right) = x e^{x/3}$

Step 5: Integrate Both Sides

Now, integrate both sides with respect to $x$:

$3u e^{x/3} = \int x e^{x/3} dx$

To solve this integral, we use integration by parts. Let $v = x$ and $dw = e^{x/3} dx$. After solving the integral, we get:

$\int x e^{x/3} dx = 3(x - 3)e^{x/3} + C$

Thus:

$3u e^{x/3} = 3(x - 3)e^{x/3} + C$

Step 6: Solve for $u$

Divide through by $3e^{x/3}$:

$u = x - 3 + \frac{C}{3} e^{-x/3}$

Step 7: Substitute Back for $y$

Recall that $y = u^3$, so:

$y = \left( x - 3 + \frac{C}{3} e^{-x/3} \right)^3$

This is the general solution to the given differential equation.

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Let me know if you need more clarification on any steps!

Related Questions:

1. How can we verify the solution to a differential equation?
2. What other methods can solve non-linear differential equations?
3. Can you solve the equation using a numerical method like Euler’s method?
4. How would the solution change if an initial condition were provided?
5. What is the role of the integrating factor in linear differential equations?

Tip:

When solving differential equations, substitutions often simplify non-linear terms, transforming them into solvable forms.