Write integrating factor differential equations
`x^2 (dy)/(dx)+y= x^4`

Answer

To solve the differential equation \( x^2 \frac{dy}{dx} + y = x^4 \) and find the integrating factor, we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ x^2 \frac{dy}{dx} + y = x^4 \]
To put it in standard linear form, we divide the entire equation by \( x^2 \):
\[ \frac{dy}{dx} + \frac{1}{x^2} y = x^2 \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
In the standard form \( \frac{dy}{dx} + p(x) y = q(x) \), we identify:
- \( p(x) = \frac{1}{x^2} \)
- \( q(x) = x^2 \)

### Step 3: Calculate the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} \]
Substituting \( p(x) = \frac{1}{x^2} \):
\[ \mu(x) = e^{\int \frac{1}{x^2} \, dx} \]

### Step 4: Solve the integral
The integral \( \int \frac{1}{x^2} \, dx \) can be computed as follows:
\[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -\frac{1}{x} + C \]
Thus, the integrating factor becomes:
\[ \mu(x) = e^{-\frac{1}{x} + C} = e^{-\frac{1}{x}} \cdot e^C \]
Since \( e^C \) is a constant, we can ignore it for the purpose of finding the integrating factor. Therefore:
\[ \mu(x) = e^{-\frac{1}{x}} \]

### Step 5: Conclusion
The integrating factor for the differential equation \( x^2 \frac{dy}{dx} + y = x^4 \) is:
\[ \mu(x) = e^{-\frac{1}{x}} \]