Find the general solution of the differential equation.​ Then, use the initial condition to find the corresponding particular solution. xy'+3y=4x, y(1)=6

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:

\[
y = x + \frac{5}{x^3}
\]

---

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.

Find the general solution of the differential equation. Then, use the initial condition to find the corresponding particular solution. xy'+3y=4x, y(1)=6

Learn how to solve the first-order linear differential equation xy'+3y=4x and find the particular solution using the initial condition y(1)=6. This problem involves applying the integrating factor method and solving step-by-step.

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