We are given the first-order linear differential equation:

$xy' + 3y = 4x$

and the initial condition $y(1) = 6$.

Step 1: Rewrite the equation in standard form

First, we divide the entire equation by $x$ to rewrite it in standard form:

$y' + \frac{3}{x} y = 4$

This is a linear first-order differential equation of the form $y' + P(x)y = Q(x)$, where:

• $P(x) = \frac{3}{x}$
• $Q(x) = 4$

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3 \ln |x|} = |x|^3$

Since $x > 0$, we can simplify $\mu(x) = x^3$.

Step 3: Multiply the equation by the integrating factor

We now multiply the entire equation by the integrating factor $x^3$:

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

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

$\frac{d}{dx} (x^3 y) = 4x^3$

Step 4: Integrate both sides

Integrating both sides with respect to $x$:

$x^3 y = \int 4x^3 \, dx = x^4 + C$

where $C$ is the constant of integration. Thus, the general solution is:

$y = \frac{x^4 + C}{x^3} = x + \frac{C}{x^3}$

Step 5: Apply the initial condition

We are given $y(1) = 6$. Substituting $x = 1$ and $y = 6$ into the general solution:

$6 = 1 + \frac{C}{1^3}$

Solving for $C$:

$6 = 1 + C \quad \Rightarrow \quad C = 5$

Step 6: Write the particular solution

Substitute $C = 5$ into the general solution to get the particular solution:

$y = x + \frac{5}{x^3}$

Thus, the particular solution is:

$y = x + \frac{5}{x^3}$

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Would you like to explore any of the steps in more detail, or do you have any questions?

Here are 5 related questions for further practice:

1. Solve $xy' + 2y = 3x^2$ with $y(1) = 4$.
2. Solve $y' - \frac{2}{x} y = x^2$ with $y(2) = 1$.
3. Find the general solution of $y' + y = x^2$.
4. Solve the differential equation $x^2 y' + 4xy = x^5$.
5. Solve $y' + 5y = 10x$ with the initial condition $y(0) = 2$.

Tip: For linear first-order differential equations, always look for an integrating factor to simplify the equation.