Questions: Find the solution of the given initial value problem in explicit form. Determine the interval in which the solution is defined. y' = 4x / (1+2y), y(2)=0 y(x)= The solution is defined on the interval

Find the solution of the given initial value problem in explicit form. Determine the interval in which the solution is defined.

y' = 4x / (1+2y), y(2)=0
y(x)=

The solution is defined on the interval

Transcript text: Find the solution of the given initial value problem in explicit form. Determine the interval in which the solution is defined. \[ \begin{array}{l} y^{\prime}=\frac{4 x}{1+2 y}, \quad y(2)=0 \\ y(x)=\square \end{array} \] $\square$ The solution is defined on the interval $(\square, \infty)$

Solution

Solution Steps

To solve the given initial value problem, we need to follow these steps:

Separate the variables $ y $ and $ x $ to opposite sides of the equation.
Integrate both sides to find the general solution.
Apply the initial condition to determine the constant of integration.
Solve for $ y $ explicitly.
Determine the interval in which the solution is defined.

Step 1: Separate the Variables

Given the differential equation: \[ y' = \frac{4x}{1 + 2y} \] We separate the variables $ y $ and $ x $: \[ \frac{1}{1 + 2y} \, dy = 4x \, dx \]

Step 2: Integrate Both Sides

Integrate both sides: \[ \int \frac{1}{1 + 2y} \, dy = \int 4x \, dx \] The left-hand side integrates to: \[ \frac{1}{2} \ln |1 + 2y| \] The right-hand side integrates to: \[ 2x^2 \] Thus, the general solution is: \[ \frac{1}{2} \ln |1 + 2y| = 2x^2 + C \]

Step 3: Apply the Initial Condition

Given the initial condition $ y(2) = 0 $, substitute $ x = 2 $ and $ y = 0 $ into the general solution: \[ \frac{1}{2} \ln |1 + 2(0)| = 2(2)^2 + C \] \[ 0 = 8 + C \] Solving for $ C $: \[ C = -8 \]

Step 4: Solve for $ y $ Explicitly

Substitute $ C = -8 $ back into the general solution: \[ \frac{1}{2} \ln |1 + 2y| = 2x^2 - 8 \] Multiply both sides by 2: \[ \ln |1 + 2y| = 4x^2 - 16 \] Exponentiate both sides to solve for $ y $: \[ |1 + 2y| = e^{4x^2 - 16} \] Since $ 1 + 2y $ must be positive: \[ 1 + 2y = e^{4x^2 - 16} \] Solve for $ y $: \[ 2y = e^{4x^2 - 16} - 1 \] \[ y = \frac{e^{4x^2 - 16} - 1}{2} \]

Step 5: Determine the Interval of Definition

The solution is valid as long as the denominator $ 1 + 2y $ is not zero. Since $ e^{4x^2 - 16} $ is always positive, $ 1 + 2y $ will never be zero. Therefore, the solution is defined for all $ x $.