Let's solve question 23:

Given the differential equation:

$\frac{dy}{dx} = \frac{x^3}{y}$

with the initial condition $y(0) = 5$.

Step 1: Separate Variables

Separate $x$ and $y$ to integrate each side independently.

$y \, dy = x^3 \, dx$

Step 2: Integrate Both Sides

Now integrate both sides with respect to their respective variables.

$\int y \, dy = \int x^3 \, dx$

The left side integrates to:

$\int y \, dy = \frac{y^2}{2}$

The right side integrates to:

$\int x^3 \, dx = \frac{x^4}{4}$

So we have:

$\frac{y^2}{2} = \frac{x^4}{4} + C$

Step 3: Multiply Both Sides by 2 to Simplify

$y^2 = \frac{x^4}{2} + 2C$

Let $2C = C'$, where $C'$ is a new constant, giving us:

$y^2 = \frac{x^4}{2} + C'$

Step 4: Apply the Initial Condition

Using the initial condition $y(0) = 5$:

$5^2 = \frac{0^4}{2} + C'$

$25 = C'$

So the particular solution becomes:

$y^2 = \frac{x^4}{2} + 25$

Step 5: Solve for $y$

To get $y$ explicitly, take the square root of both sides:

$y = \pm \sqrt{\frac{x^4}{2} + 25}$

Given $y(0) = 5$, we select the positive branch:

$y = \sqrt{\frac{x^4}{2} + 25}$

Final Solution

The solution to the differential equation is:

$y = \sqrt{\frac{x^4}{2} + 25}$

Would you like further details on any of the steps, or have any questions?

Here are some related questions that might help deepen your understanding:

1. How would the solution change if the initial condition were different?
2. What happens if we select the negative branch of the square root?
3. Can this technique be applied to other differential equations of similar form?
4. What is the significance of separating variables in differential equations?
5. How does the initial condition affect the constant of integration?

Tip: Always check if you can separate variables directly when solving differential equations, as it simplifies the process considerably!