Solve xdydx=2y−6x,y(1)=8
.

(a) Identify the integrating factor, r(x)
.
r(x)
 = 
x^(-2)
(b) Find the general solution.
y(x)=
 
6x+2x^2

Note: Use C
 for the arbitrary constant.

(c) Solve the initial value problem y(1)=8
.
y(x)=
 


Let's solve the given differential equation step by step.

### Given:
\[
x \frac{dy}{dx} = 2y - 6x, \quad y(1) = 8
\]

### Step 1: Rewrite the equation in standard linear form.
The standard linear form for a first-order differential equation is:
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
First, divide the equation by \(x\):
\[
\frac{dy}{dx} - \frac{2}{x}y = -6
\]

Here, \(P(x) = -\frac{2}{x}\) and \(Q(x) = -6\).

### Step 2: Identify the integrating factor \(r(x)\).
The integrating factor \(r(x)\) is given by:
\[
r(x) = e^{\int P(x) \, dx} = e^{\int -\frac{2}{x} \, dx}
\]
\[
r(x) = e^{-2 \ln|x|} = e^{\ln|x^{-2}|} = |x|^{-2} = \frac{1}{x^2}
\]

So, \(r(x) = \frac{1}{x^2}\).

### Step 3: Multiply the equation by the integrating factor.
Multiply the entire differential equation by \(\frac{1}{x^2}\):
\[
\frac{1}{x^2} \frac{dy}{dx} - \frac{2}{x^3}y = \frac{-6}{x^2}
\]
This simplifies to:
\[
\frac{d}{dx} \left( \frac{y}{x^2} \right) = \frac{-6}{x^2}
\]

### Step 4: Integrate both sides.
Integrate both sides with respect to \(x\):
\[
\frac{y}{x^2} = \int \frac{-6}{x^2} \, dx
\]
\[
\frac{y}{x^2} = 6x^{-1} + C
\]
\[
\frac{y}{x^2} = \frac{6}{x} + C
\]

### Step 5: Solve for \(y(x)\).
Multiply through by \(x^2\) to solve for \(y(x)\):
\[
y(x) = 6x + Cx^2
\]

### Step 6: Apply the initial condition \(y(1) = 8\).
Substitute \(x = 1\) and \(y(1) = 8\) into the general solution:
\[
8 = 6(1) + C(1)^2
\]
\[
8 = 6 + C
\]
\[
C = 2
\]

### Final Solution:
Substituting \(C = 2\) into the general solution, we get:
\[
y(x) = 6x + 2x^2
\]

### Answer Summary:
- **(a)** The integrating factor \(r(x) = \frac{1}{x^2}\).
- **(b)** The general solution \(y(x) = 6x + Cx^2\).
- **(c)** The solution to the initial value problem \(y(x) = 6x + 2x^2\).

Would you like more details or have any other questions?

Here are five related questions you might find interesting:
1. How would the solution change if the initial condition was different?
2. Can you explain the importance of the integrating factor in solving linear differential equations?
3. How can you verify that the solution satisfies the original differential equation?
4. What are some other methods to solve first-order linear differential equations?
5. What happens if you have a non-linear first-order differential equation?

**Tip:** The method of integrating factors is powerful for solving linear first-order differential equations, especially when the equation is not exact.

Learn how to solve the first-order linear differential equation x dy/dx = 2y - 6x with the initial condition y(1) = 8. This problem involves finding the integrating factor, deriving the general solution, and solving the initial value problem. Suitable for advanced learners and those studying differential equations.

Math Problem Statement

(a) Identify the integrating factor, r(x) . r(x)

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