Question

Solve the initial value problem and graph the solution. $$ y^{\prime}=2 x y\left(1+y^{2}\right), \quad y(0)=1 $$

Short answer

To summarize, the initial value problem (IVP) for the differential equation \(y'=2xy(1+y^2)\) with the initial condition \(y(0)=1\) was solved using the following steps: 1. Separate variables: Rewrote the equation as \(\frac{dy}{1+y^2(1+y^2)} = 2x \cdot dx\) 2. Integrate both sides: Obtained \(A = \frac{1}{2}\left[\ln \left|\frac{1+y^2}{y^2}\right|\right]\) and \(B = x^2 + C\) 3. Apply initial condition: Found the particular solution: \(\frac{1}{2}\left[\ln \left|\frac{1+y^2}{y^2}\right|\right] = x^2 + \ln(2)\) 4. Graph the solution: The graph of the particular solution can be plotted using graphing software, and it forms a curve that satisfies the given initial condition \((0, 1)\).

Step-by-step solution

Separate Variables

Separate the variables in the given differential equation, moving the \(x\) terms to one side and the \(y\) terms to the other side. The result is: $$ \frac{dy}{1+y^2(1+y^2)} = 2x \cdot dx $$

Integrate Both Sides

Integrate both sides of the separated equation: $$ \int \frac{dy}{1+y^2(1+y^2)} = \int 2x \cdot dx $$ Let's call the left side integral \(A\) and the right side integral \(B\). Now we need to find the values of \(A\) and \(B\). For \(B\), it's pretty simple: $$ B = \int 2x\cdot dx = x^2 + C $$ For \(A\), we can use the substitution method. Let \(u=1+y^2\), then \(du=2y\cdot dy\). So our integral \(A\) becomes: $$ A = \int \frac{1}{u(u-1)} \cdot \frac{du}{2} = \frac{1}{2}\int \frac{1}{u(u-1)}\cdot du $$ Now, in order to compute the integral \(A\), we can use partial fraction decomposition: $$ \frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1} $$ Multiplying both sides by \(u(u-1)\), we get: $$ 1 = A(u-1) + B \cdot u $$ To find \(A\) and \(B\), we can solve this system of equations: $$ \begin{cases} (1) \quad 1=A(1) + B(1) \\ (2) \quad 0=A(0) + B(-1) \end{cases} $$ The result is: \(A=1\) and \(B=-1\). Replacing these values in the partial fraction decomposition: $$ \frac{1}{u(u-1)} = \frac{1}{u} - \frac{1}{u-1} $$ Therefore, we can write our integral \(A\) as: $$ A = \frac{1}{2}\int \left( \frac{1}{u} - \frac{1}{u-1} \right)\cdot du = \frac{1}{2}\left[\ln \left|\frac{u}{u-1}\right|\right] $$ Now, we can replace \(u\) with its value in terms of \(y\): $$ A = \frac{1}{2}\left[\ln \left|\frac{1+y^2}{y^2}\right|\right] $$

Apply Initial Condition and Solve for \(C\)

We have \(A = x^2 + C\), and we can write the solution as: $$ \frac{1}{2}\left[\ln \left|\frac{1+y^2}{y^2}\right|\right] = x^2 + C $$ Now, apply the initial condition \(y(0) = 1\): $$ \frac{1}{2}\left[\ln \left|\frac{1+1^2}{1^2}\right|\right] = (0)^2 + C $$ So, \(C = \ln(2)\). Our particular solution becomes: $$ \frac{1}{2}\left[\ln \left|\frac{1+y^2}{y^2}\right|\right] = x^2 + \ln(2) $$

Graph the Solution

Although graphing this particular solution by hand can be challenging due to the implicit form, you can plot the solution using graphing software like Desmos, GeoGebra, or Wolfram Alpha. It would appear as a curve, with the given initial condition \((0, 1)\).

Key concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are crucial in modeling various real-world phenomena, including population dynamics, physics, and engineering problems. The differential equation in this exercise is given as:
\[ y^{\prime}=2 x y\left(1+y^{2}\right) \]This equation is a first-order differential equation because it involves the first derivative of the function \( y \). The solution to a differential equation is a function, or a set of functions, that satisfies the equation. Often, additional information, such as initial conditions, is needed to find a unique solution. In our exercise, the initial condition is \( y(0)=1 \), helping us to determine the specific function \( y \) that satisfies both the equation and this initial condition.

Variable Separation

Separation of variables is a method used to simplify differential equations, allowing them to be more easily solved. The idea is to separate the variables, in this case \( x \) and \( y \), on different sides of the equation.
In our example, we start with the equation:
\[ y^{\prime}=2 x y\left(1+y^{2}\right) \]By isolating \( dy \) and \( dx \) on opposite sides:
\[ \frac{dy}{1+y^2(1+y^2)} = 2x \cdot dx \]We intersperse \( dy \) with terms containing only \( y \) and \( dx \) with terms containing only \( x \), allowing us to integrate both sides separately. This step makes the differential equation more manageable and solvable.

Integration Methods

Integration is the key technique used after separating variables. The goal is to solve for the specific solution that satisfies the given condition. Each side of the separated equation needs to be integrated:

• Left side: \( \int \frac{dy}{1+y^2(1+y^2)} \)
• Right side: \( \int 2x \cdot dx \)

For the right side, the integral \( B \) is straightforward:
\[ B = x^2 + C \]For the left side integral \( A \), use substitution; set \( u=1+y^2 \), \( du=2y \cdot dy \). This transforms \( A \) into a solvable form using partial fraction decomposition:
\[ A = \frac{1}{2}\left[\ln \left|\frac{u}{u-1}\right|\right] \]Remember to substitute \( u \) back to \( 1+y^2 \) at the end.

Partial Fraction Decomposition

Partial fraction decomposition is used when integrating complex rational expressions, such as those encountered in our differential equation. It helps decompose a fraction into simpler parts that are easier to integrate.
Given:
\[ \frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1} \]Multiply through by \( u(u-1) \) to clear the denominators:
\[ 1 = A(u-1) + B \cdot u \]By comparing coefficients, solve to find \( A = 1 \) and \( B = -1 \). These values are then substituted back into the partial fraction equation, simplifying it for integration:
\[ A = \frac{1}{2}\int \left( \frac{1}{u} - \frac{1}{u-1} \right)\cdot du \]Solving this provides the setup for completing the integral on the left side of the separated variables equation.