Find the general solution of :
`(dy)/(dx)=x^(2)+sin 3x`.

Answer

To find the general solution of the differential equation

\[ \frac{dy}{dx} = x^2 + \sin(3x), \]

we will integrate both sides with respect to \(x\).

### Step 1: Integrate both sides

We start with the equation:

\[ \frac{dy}{dx} = x^2 + \sin(3x). \]

Integrating both sides with respect to \(x\):

\[ \int \frac{dy}{dx} \, dx = \int (x^2 + \sin(3x)) \, dx. \]

### Step 2: Solve the left side

The left side simplifies to:

\[ y = \int (x^2 + \sin(3x)) \, dx. \]

### Step 3: Integrate the right side

Now we will integrate the right side term by term:

1. **Integrate \(x^2\)**:
\[ \int x^2 \, dx = \frac{x^3}{3}. \]

2. **Integrate \(\sin(3x)\)**:
To integrate \(\sin(3x)\), we use the substitution:
\[ u = 3x \quad \Rightarrow \quad du = 3 \, dx \quad \Rightarrow \quad dx = \frac{du}{3}. \]
Thus,
\[ \int \sin(3x) \, dx = \int \sin(u) \cdot \frac{du}{3} = \frac{1}{3} \int \sin(u) \, du = -\frac{1}{3} \cos(u) + C = -\frac{1}{3} \cos(3x). \]

### Step 4: Combine the results

Combining both integrals, we have:

\[ y = \frac{x^3}{3} - \frac{1}{3} \cos(3x) + C, \]

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

### Final Solution

Thus, the general solution of the differential equation is:

\[ y = \frac{x^3}{3} - \frac{1}{3} \cos(3x) + C. \]

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