Math Problem Statement

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

Solution

To solve the differential equation 2yy=x2yy' = x with the initial condition y(2)=3y(2) = 3, follow these steps:

Step 1: Separate the variables

The given differential equation is:

2yy=x2yy' = x

First, rewrite yy' as dydx\frac{dy}{dx}:

2ydydx=x2y \frac{dy}{dx} = x

Now, separate the variables to get all terms involving yy on one side and terms involving xx on the other:

2ydy=xdx2y \, dy = x \, dx

Step 2: Integrate both sides

Integrate both sides:

2ydy=xdx\int 2y \, dy = \int x \, dx

The integrals are straightforward:

y2=x22+Cy^2 = \frac{x^2}{2} + C

Step 3: Solve for yy

Take the square root of both sides:

y=±x22+Cy = \pm \sqrt{\frac{x^2}{2} + C}

Step 4: Apply the initial condition

We are given that y(2)=3y(2) = 3. Substituting x=2x = 2 and y=3y = 3 into the equation:

3=222+C3 = \sqrt{\frac{2^2}{2} + C}

3=42+C3 = \sqrt{\frac{4}{2} + C}

3=2+C3 = \sqrt{2 + C}

Square both sides to eliminate the square root:

9=2+C9 = 2 + C

Solve for CC:

C=7C = 7

Step 5: Final solution

Substitute C=7C = 7 into the equation for yy:

y=±x22+7y = \pm \sqrt{\frac{x^2}{2} + 7}

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

y(x)=x22+7y(x) = \sqrt{\frac{x^2}{2} + 7}

Final answer:

y(x)=x22+7y(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.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Separation of Variables
Initial Conditions

Formulas

2yy' = x
y = √((x²/2) + C)

Theorems

Fundamental Theorem of Calculus
Separable Differential Equations

Suitable Grade Level

Undergraduate Math (Calculus I/II)