Solve the differential equation 
2yy' + 3x = 0
where x =1, and y=1

To solve the differential equation:

\[
2yy' + 3x = 0
\]

where \( y' = \frac{dy}{dx} \), let's first rewrite the equation in a more standard form:

\[
2y \frac{dy}{dx} = -3x
\]

Now, separate the variables:

\[
2y \, dy = -3x \, dx
\]

Integrate both sides:

\[
\int 2y \, dy = \int -3x \, dx
\]

This gives:

\[
y^2 = -\frac{3}{2}x^2 + C
\]

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

### Applying the Initial Conditions

We are given \( x = 1 \) and \( y = 1 \). Substitute these values into the equation to solve for \( C \):

\[
(1)^2 = -\frac{3}{2}(1)^2 + C
\]

Simplify:

\[
1 = -\frac{3}{2} + C
\]

\[
C = 1 + \frac{3}{2} = \frac{5}{2}
\]

### The General Solution

Substitute \( C = \frac{5}{2} \) back into the equation:

\[
y^2 = -\frac{3}{2}x^2 + \frac{5}{2}
\]

To express \( y \) explicitly, take the square root:

\[
y = \pm \sqrt{-\frac{3}{2}x^2 + \frac{5}{2}}
\]

### Selecting the Correct Solution

Since \( y = 1 \) when \( x = 1 \), we choose the positive square root:

\[
y = \sqrt{-\frac{3}{2}x^2 + \frac{5}{2}}
\]

### Final Solution

Thus, the solution to the differential equation is:

\[
\boxed{y = \sqrt{-\frac{3}{2}x^2 + \frac{5}{2}}}
\]

Would you like further details, or do you have any questions?

### Related Questions:

1. How would the solution change if different initial conditions were given?
2. How can we verify if a function is a solution to a differential equation?
3. What are the implications of the constant of integration in such problems?
4. What are the methods to solve non-separable differential equations?
5. How do different forms of differential equations impact their solutions?

**Tip:** Always check your solutions by substituting them back into the original differential equation to ensure they satisfy it.

Solve the differential equation 2yy' + 3x = 0 where x =1, and y=1.

Explore how to solve the differential equation 2yy' + 3x = 0 using separation of variables and integration techniques. Learn the step-by-step process to find the solution with given initial conditions (x=1, y=1) for advanced high school or undergraduate calculus students.

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