Let's solve the problem shown in the image.

We are asked to use separation of variables to solve the following initial value problem:

$\frac{dy}{dx} = 2x \cos 2x \quad \text{with} \quad y(3) = 0$

Step 1: Separate the variables

Rewriting the equation:

$\frac{dy}{dx} = 2x \cos(2x)$

Multiply both sides by $dx$ to separate $y$ and $x$:

$dy = 2x \cos(2x) \, dx$

Step 2: Integrate both sides

We now integrate both sides. Starting with the right-hand side:

$\int dy = \int 2x \cos(2x) \, dx$

To solve the integral on the right, we can use integration by parts. Let:

• $u = x$, so $du = dx$
• $dv = 2 \cos(2x) dx$, so $v = \sin(2x)$

Using the formula for integration by parts:

$\int u \, dv = uv - \int v \, du$

We get:

$\int 2x \cos(2x) \, dx = 2 \left( x \sin(2x) - \int \sin(2x) \, dx \right)$

Now, integrate $\int \sin(2x) dx$:

$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)$

Thus, the integral becomes:

$2 \left( x \sin(2x) + \frac{1}{2} \cos(2x) \right)$

Simplifying:

$2x \sin(2x) + \cos(2x)$

Step 3: Write the general solution

Now, the equation for $y(x)$ becomes:

$y = 2x \sin(2x) + \cos(2x) + C$

Step 4: Use the initial condition

We are given that $y(3) = 0$. Substitute $x = 3$ and $y = 0$ to find $C$:

$0 = 2(3) \sin(6) + \cos(6) + C$

$0 = 6 \sin(6) + \cos(6) + C$

Thus,

$C = -6 \sin(6) - \cos(6)$

Final solution:

$y = 2x \sin(2x) + \cos(2x) - 6 \sin(6) - \cos(6)$

Would you like more details on any specific part? Here are some related questions:

1. How does integration by parts work in detail?
2. Why did we choose to use integration by parts in this case?
3. Can we graph this solution for different values of $x$?
4. How do we verify the solution with the initial condition?
5. Are there alternative methods to solve this differential equation?

Tip: When solving differential equations, always check if integration by parts is the simplest method, or if there's a substitution that might simplify it further.