Solve the following differential equations :
`(dy)/(dx)-y=3x^(3)`.

Answer

To solve the differential equation

\[ \frac{dy}{dx} - y = 3x^3, \]

we will follow these steps:

### Step 1: Identify the Standard Form
The given equation can be rewritten in standard linear form:

\[ \frac{dy}{dx} + P(x)y = Q(x), \]

where \( P(x) = -1 \) and \( Q(x) = 3x^3 \).

### Step 2: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -1 \, dx} = e^{-x}. \]

### Step 3: Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by the integrating factor:

\[ e^{-x} \frac{dy}{dx} - e^{-x}y = 3x^3 e^{-x}. \]

This simplifies to:

\[ \frac{d}{dx}(y e^{-x}) = 3x^3 e^{-x}. \]

### Step 4: Integrate Both Sides
Now, integrate both sides with respect to \( x \):

\[ \int \frac{d}{dx}(y e^{-x}) \, dx = \int 3x^3 e^{-x} \, dx. \]

The left side simplifies to:

\[ y e^{-x} = \int 3x^3 e^{-x} \, dx + C, \]

where \( C \) is the constant of integration.

### Step 5: Solve the Integral on the Right Side
To solve the integral \( \int 3x^3 e^{-x} \, dx \), we can use integration by parts. Let:

- \( u = x^3 \) and \( dv = 3e^{-x}dx \)
- Then \( du = 3x^2dx \) and \( v = -3e^{-x} \)

Applying integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ \int 3x^3 e^{-x} \, dx = -3x^3 e^{-x} - \int -3e^{-x} (3x^2) \, dx. \]

This leads to:

\[ \int 3x^3 e^{-x} \, dx = -3x^3 e^{-x} + 9 \int x^2 e^{-x} \, dx. \]

We can repeat the integration by parts for \( \int x^2 e^{-x} \, dx \) and continue this process until we reach a solvable integral.

### Step 6: Substitute Back
After calculating the integrals, we substitute back into the equation:

\[ y e^{-x} = \text{(result of integration)} + C. \]

### Step 7: Solve for \( y \)
Finally, solve for \( y \):

\[ y = e^{x} \left( \text{(result of integration)} + C \right). \]

### Final Solution
The final solution will be in the form:

\[ y = e^{x} \left( -3x^3 + 9x^2 - 18x + 18 + C \right). \]