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 \(x^3\).