Math Problem Statement

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

Solution

To solve the differential equation 2ydx=3xdy2y \, dx = 3x \, dy using the method of separation of variables and obtain the particular solution given the initial condition (x,y)=(2,1)(x, y) = (2, 1), follow these steps:

Step 1: Rewrite the differential equation

The given differential equation is:

2ydx=3xdy2y \, dx = 3x \, dy

Divide both sides by ydxy \, dx (assuming y0y \neq 0 and dx0dx \neq 0) to separate variables:

23xdx=1ydy\frac{2}{3x} \, dx = \frac{1}{y} \, dy

Step 2: Integrate both sides

Integrate both sides with respect to their respective variables:

23xdx=1ydy\int \frac{2}{3x} \, dx = \int \frac{1}{y} \, dy

The integrals are:

231xdx=1ydy\frac{2}{3} \int \frac{1}{x} \, dx = \int \frac{1}{y} \, dy

23lnx=lny+C\frac{2}{3} \ln |x| = \ln |y| + C

Step 3: Solve for yy

Exponentiate both sides to remove the logarithm:

e23lnx=elny+Ce^{\frac{2}{3} \ln |x|} = e^{\ln |y| + C}

This simplifies to:

x23=yeCx^{\frac{2}{3}} = y e^C

Let eC=Ke^C = K (where KK is a constant):

x23=Kyx^{\frac{2}{3}} = Ky

Solve for yy:

y=x23Ky = \frac{x^{\frac{2}{3}}}{K}

Step 4: Apply the initial condition

Given y(2)=1y(2) = 1, substitute x=2x = 2 and y=1y = 1 into the equation:

1=223K1 = \frac{2^{\frac{2}{3}}}{K}

Solve for KK:

K=223K = 2^{\frac{2}{3}}

Step 5: Write the particular solution

Substitute KK back into the equation for yy:

y=x23223y = \frac{x^{\frac{2}{3}}}{2^{\frac{2}{3}}}

This simplifies to:

y=(x2)23y = \left(\frac{x}{2}\right)^{\frac{2}{3}}

Thus, the particular solution is:

y=(x2)23\boxed{y = \left(\frac{x}{2}\right)^{\frac{2}{3}}}

Would you like any additional details or have any questions?


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 xx 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.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Separation of Variables
Initial Conditions

Formulas

\( \int \frac{1}{x} \, dx = \ln |x| + C \)
\( y = \left(\frac{x}{2}\right)^{\frac{2}{3}} \)

Theorems

Method of Separation of Variables

Suitable Grade Level

Grades 11-12