Write integrating factor differential equations
`x(dy)/(dx)-3y= x^3`

Answer

To solve the differential equation \( x \frac{dy}{dx} - 3y = x^3 \), we will first rewrite it in standard linear form and then find the integrating factor.

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ x \frac{dy}{dx} - 3y = x^3 \]
To put it in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), we divide the entire equation by \( x \):
\[ \frac{dy}{dx} - \frac{3}{x}y = x^2 \]

### Step 2: Identify \( P(x) \) and \( Q(x) \)
From the rewritten equation, we can identify:
\[ P(x) = -\frac{3}{x}, \quad Q(x) = x^2 \]

### Step 3: Calculate the integrating factor
The integrating factor \( IF \) is given by:
\[ IF = e^{\int P(x) \, dx} \]
Substituting \( P(x) \):
\[ IF = e^{\int -\frac{3}{x} \, dx} \]
Calculating the integral:
\[ \int -\frac{3}{x} \, dx = -3 \ln |x| = \ln |x|^{-3} \]
Thus, the integrating factor becomes:
\[ IF = e^{\ln |x|^{-3}} = |x|^{-3} \]
Since \( x \) is positive in this context, we can simplify:
\[ IF = \frac{1}{x^3} \]

### Step 4: Multiply the entire differential equation by the integrating factor
Now we multiply the entire differential equation by \( \frac{1}{x^3} \):
\[ \frac{1}{x^3} \frac{dy}{dx} - \frac{3}{x^4}y = 1 \]

### Step 5: Recognize the left-hand side as a derivative
The left-hand side can be recognized as the derivative of a product:
\[ \frac{d}{dx}\left(\frac{y}{x^3}\right) = 1 \]

### Step 6: Integrate both sides
Now we integrate both sides:
\[ \int \frac{d}{dx}\left(\frac{y}{x^3}\right) \, dx = \int 1 \, dx \]
This gives us:
\[ \frac{y}{x^3} = x + C \]
where \( C \) is the constant of integration.

### Step 7: Solve for \( y \)
Multiplying through by \( x^3 \) to solve for \( y \):
\[ y = x^4 + Cx^3 \]

### Final Solution
The general solution to the differential equation is:
\[ y = x^4 + Cx^3 \]