The given differential equation is:

$x y' = 3y - 6x^2$

Step 1: Rearranging the equation

First, let's divide both sides by $x$ (assuming $x \neq 0$) to isolate $y'$:

$y' = \frac{3y}{x} - 6x$

This is a first-order linear differential equation.

Step 2: Solving the equation

This equation is in the form $y' + P(x)y = Q(x)$, where:

• $P(x) = -\frac{3}{x}$
• $Q(x) = -6x$

We can solve it using an integrating factor.

Step 3: Finding the integrating factor

The integrating factor is given by: $\mu(x) = e^{\int P(x) dx} = e^{\int -\frac{3}{x} dx}$

$\mu(x) = e^{-3 \ln |x|} = x^{-3}$

Step 4: Multiplying both sides by the integrating factor

Now, multiply both sides of the original equation by $x^{-3}$:

$x^{-3} y' + x^{-3} \cdot \frac{3y}{x} = x^{-3} (-6x)$

Simplifying:

$x^{-3} y' + 3x^{-4} y = -6x^{-2}$

The left-hand side is now the derivative of $x^{-3} y$:

$\frac{d}{dx} (x^{-3} y) = -6x^{-2}$

Step 5: Integrating both sides

Integrate both sides with respect to $x$:

$x^{-3} y = \int -6x^{-2} dx$

$x^{-3} y = 6x^{-1} + C$

Step 6: Solving for $y$

Multiply both sides by $x^3$ to solve for $y$:

$y = 6x^2 + Cx^3$

Final Solution:

$y(x) = 6x^2 + Cx^3$ where $C$ is the constant of integration.

Would you like further details or have any questions?

Here are some related questions you can explore:

1. How do you verify the solution of a differential equation?
2. What is an integrating factor, and why is it used?
3. Can all first-order linear differential equations be solved using integrating factors?
4. How do initial conditions affect the solution of this type of differential equation?
5. What are some practical applications of first-order differential equations?

Tip: When solving linear differential equations, identifying the correct integrating factor simplifies the process significantly.