Question

Consider the initial value problem \(y^{\prime}=\sqrt{1-y^{2}}, y(0)=0\) (a) Show that \(y=\sin x\) is the solution of this initial value problem. (b) Look for a solution of the initial value problem in the form of a power series about \(x=0 .\) Find the coefficients up to the term in \(x^{3}\) in this series.

Short answer

Question: Verify that \(y=\sin x\) is a solution of the initial value problem \(y'=\sqrt{1-y^2}\) and \(y(0)=0\). Then, find a power series representation of the solution around \(x=0\) up to the term in \(x^3\). Answer: We have verified that \(y=\sin x\) is indeed a solution of the given initial value problem. The power series representation of the solution around \(x=0\) up to the term in \(x^3\) is \(y(x) = 1x - x^2\).

Step-by-step solution

Verify the given solution

To verify that \(y=\sin x\) is a solution of the IVP, we first need to compute its derivative with respect to \(x\). Using the basic rules of differentiation, we have: \(y' = \frac{d}{dx}(\sin x)=\cos x\) Next, we find \(\sqrt{1-y^2}\). Since \(y=\sin x,\) we have: \(\sqrt{1-y^2} = \sqrt{1-\sin^2 x} = \sqrt{\cos^2 x} = |\cos{x}|\) Since we are considering \(x\) around zero, and the cosine function is positive at \(x=0\), we can assume that the expression for \(|\cos x|\) is equal to \(\cos x\). Now we compare the expressions for \(y'\) and \(\sqrt{1-y^2}\): \(y' = \cos x = \sqrt{1-y^2}\) Since these expressions are equal, we can conclude that \(y=\sin x\) is indeed a solution of the given IVP.

Form the power series representation

Now we will look for a solution in the power series form: \(y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots\) We'll also need the first derivative of this expression: \(y'(x) = a_1 + 2 a_2 x + 3 a_3 x^2 + \cdots\) Now, we substitute the power series for \(y\) and \(y'\) into the given equation: \(a_1 + 2 a_2 x + 3 a_3 x^2 + \cdots =\sqrt{1-(a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots)^2}\) Now we need to find the coefficients up to the term in \(x^3\). In order to do that, we can equate the powers of \(x\) on both sides of the equation and use the initial condition: - For \(x^0\): \(a_1 = \sqrt{1-a_0^2}\) and \(y(0)=0\), so \(a_0=0\), \(a_1=1\). - For \(x^1\): \(2a_2 = -2a_1^2 = -2\), so \(a_2 = -1\). - For \(x^2\): \(3a_3 = 0\), so \(a_3 = 0\). Now we found the coefficients up to the term in \(x^3\) and the power series representation becomes: \(y(x) = 1x -x^2\) We have verified that \(y=\sin x\) is indeed the solution of the IVP, and we found the coefficients for a power series representation up to the term in \(x^3\).

Key concepts

Power Series

A power series is an infinite series in the form of \[ y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ext{...} \]Power series play a crucial role in solving differential equations, especially around a point like \( x = 0 \). When tasked with finding a solution to an initial value problem (IVP) as a power series, you start by expressing the potential solution as this infinite polynomial and need to find its coefficients.
To solve the given IVP using a power series, note the importance of the initial condition \( y(0) = 0 \), which allows the determination of the initial coefficients, specifically \( a_0 \). The next step is equating the power series development of \( y' \), the derivative of your guessed solution, with the given differential equation.
After substitution, match terms of equivalent powers of \( x \) to find the values of coefficients up to the desired power, such as \( x^3 \). This technique helps in approximating solutions when an exact closed form is not readily apparent. For this problem, the power series method confirms the solution \( y(x) = x - x^2 \) up to the \( x^3 \) term.

Differential Equations

Differential equations involve an unknown function and its derivatives. The essence is to find a function that satisfies a given relationship between the function itself and its derivatives. It’s a cornerstone of mathematical modeling in various sciences because it can describe numerous natural phenomena.
Initial value problems (IVPs) are a specific type of differential equations where the solution must satisfy certain initial conditions. For instance, the IVP \( y' = \sqrt{1-y^2} \) with \( y(0) = 0 \) requires not just finding a function that differentiates correctly but also one that satisfies the condition at \( x = 0 \).
The beauty of differential equations lies in their application, from simple growth models like population dynamics to complex systems like fluid mechanics. In our case, using \( y = \sin x \) as a guessed solution, we verified that it satisfies both the differential equation and the initial condition, showing it as a valid solution.

Verification of Solutions

Verification of solutions is a crucial step in solving differential equations. It ensures that your proposed solution is indeed correct. For an initial value problem (IVP), verifying involves checking two main aspects:

• Whether the solution satisfies the differential equation itself.
• Whether it meets the initial condition(s) provided.

To verify that \( y = \sin x \) is indeed a solution to the IVP, we first compute its derivative, \( y' = \cos x \). We compare this with \( \sqrt{1 - y^2} \), which simplifies to \( \cos x \) given that \( y = \sin x \).
Both sides equate, confirming it solves the differential equation. Additionally, we checked the initial condition \( y(0) = 0 \), and since \( \sin(0) = 0 \), this condition holds true.
Verification confirms the validity and correctness of solutions and provides confidence in the results derived via methods like guessing or using series expansions.