Obtain the particular solution using variable separable in 2ydx = 3xdy @ (2,1) in form of y

To solve the differential equation \( 2y \, dx = 3x \, dy \) using the method of separation of variables and obtain the particular solution given the initial condition \( (x, y) = (2, 1) \), follow these steps:

### Step 1: Rewrite the differential equation
The given differential equation is:

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

Divide both sides by \( y \, dx \) (assuming \( y \neq 0 \) and \( dx \neq 0 \)) to separate variables:

\[
\frac{2}{3x} \, dx = \frac{1}{y} \, dy
\]

### Step 2: Integrate both sides
Integrate both sides with respect to their respective variables:

\[
\int \frac{2}{3x} \, dx = \int \frac{1}{y} \, dy
\]

The integrals are:

\[
\frac{2}{3} \int \frac{1}{x} \, dx = \int \frac{1}{y} \, dy
\]

\[
\frac{2}{3} \ln |x| = \ln |y| + C
\]

### Step 3: Solve for \( y \)
Exponentiate both sides to remove the logarithm:

\[
e^{\frac{2}{3} \ln |x|} = e^{\ln |y| + C}
\]

This simplifies to:

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

Let \( e^C = K \) (where \( K \) is a constant):

\[
x^{\frac{2}{3}} = Ky
\]

Solve for \( y \):

\[
y = \frac{x^{\frac{2}{3}}}{K}
\]

### Step 4: Apply the initial condition
Given \( y(2) = 1 \), substitute \( x = 2 \) and \( y = 1 \) into the equation:

\[
1 = \frac{2^{\frac{2}{3}}}{K}
\]

Solve for \( K \):

\[
K = 2^{\frac{2}{3}}
\]

### Step 5: Write the particular solution
Substitute \( K \) back into the equation for \( y \):

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

This simplifies to:

\[
y = \left(\frac{x}{2}\right)^{\frac{2}{3}}
\]

Thus, the particular solution is:

\[
\boxed{y = \left(\frac{x}{2}\right)^{\frac{2}{3}}}
\]

Would you like any additional details or have any questions?

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Here are five related questions you might find useful:
1. How would the solution change if the initial condition were different?
2. Can you explain the method of separation of variables in more detail?
3. How do you interpret the constants after integrating differential equations?
4. What are other methods to solve first-order differential equations?
5. How does the solution behave as \( x \) approaches 0 or infinity?

**Tip:** Always check the domain of the solution, especially when dealing with logarithmic or power functions, to ensure they are valid for the given conditions.

Learn how to solve the differential equation 2y dx = 3x dy using the method of separation of variables, including step-by-step integration and applying initial conditions. Ideal for students in Grades 11-12.

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