find  a function y(x) such that 2yy'=x and y(2)=3

To solve the differential equation \( 2yy' = x \) with the initial condition \( y(2) = 3 \), follow these steps:

### Step 1: Separate the variables
The given differential equation is:

\[
2yy' = x
\]

First, rewrite \( y' \) as \( \frac{dy}{dx} \):

\[
2y \frac{dy}{dx} = x
\]

Now, separate the variables to get all terms involving \( y \) on one side and terms involving \( x \) on the other:

\[
2y \, dy = x \, dx
\]

### Step 2: Integrate both sides
Integrate both sides:

\[
\int 2y \, dy = \int x \, dx
\]

The integrals are straightforward:

\[
y^2 = \frac{x^2}{2} + C
\]

### Step 3: Solve for \( y \)
Take the square root of both sides:

\[
y = \pm \sqrt{\frac{x^2}{2} + C}
\]

### Step 4: Apply the initial condition
We are given that \( y(2) = 3 \). Substituting \( x = 2 \) and \( y = 3 \) into the equation:

\[
3 = \sqrt{\frac{2^2}{2} + C}
\]

\[
3 = \sqrt{\frac{4}{2} + C}
\]

\[
3 = \sqrt{2 + C}
\]

Square both sides to eliminate the square root:

\[
9 = 2 + C
\]

Solve for \( C \):

\[
C = 7
\]

### Step 5: Final solution
Substitute \( C = 7 \) into the equation for \( y \):

\[
y = \pm \sqrt{\frac{x^2}{2} + 7}
\]

Since \( y(2) = 3 \), we choose the positive square root. Therefore, the solution is:

\[
y(x) = \sqrt{\frac{x^2}{2} + 7}
\]

### Final answer:
\[
y(x) = \sqrt{\frac{x^2}{2} + 7}
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you solve separable differential equations in general?
2. Can you explain the method of verifying a solution to a differential equation?
3. How do initial conditions affect the choice of the constant in solutions?
4. What is the difference between explicit and implicit solutions to differential equations?
5. How does the solution change if the initial condition were different?

**Tip:** Always check if a solution satisfies both the differential equation and the initial conditions when solving.

Find a function y(x) such that 2yy' = x and y(2) = 3.

Learn how to solve the differential equation 2yy' = x with the initial condition y(2) = 3. This problem covers the separation of variables technique, the integration process, and applying initial conditions. Suitable for undergraduate students studying calculus.

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