Find a power series solution of the differential equation given below. Determine the radius of convergence of the resulting series, and use the series given below to identify the series in terms of familiar elementary functions.
\[
y^{\prime}+14 x y=0
\]
\# Click the icon to view power series representations of elementary functions.
The power series solution is $y(x)=\square+\cdots$
(Type an expression in terms of $\mathrm{c}_{0}$ that includes all terms up to order 6.)

Step 1 ：The given differential equation is a first order linear homogeneous differential equation. We can solve it using power series method. The power series solution of a differential equation is a solution that is represented as an infinite sum of terms. The general form of a power series is: \[y(x) = \sum_{n=0}^{\infty} c_n x^n\] where \(c_n\) are the coefficients of the series and \(x\) is the variable. The derivative of \(y(x)\) is: \[y'(x) = \sum_{n=1}^{\infty} n c_n x^{n-1}\]

Step 2 ：Substituting these into the differential equation, we get: \[\sum_{n=1}^{\infty} n c_n x^{n-1} + 14x \sum_{n=0}^{\infty} c_n x^n = 0\] We can simplify this equation by shifting the index of the first sum and combining the two sums. Then we can solve for the coefficients \(c_n\).

Step 3 ：The solution of the differential equation is given by the expression \(c(x) = C_1 e^{-7x^2}\). However, we need to find the power series representation of this solution. We can do this by expanding the exponential function into a power series. The power series expansion of \(e^x\) is given by: \[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\] We can use this to expand \(e^{-7x^2}\) into a power series. Then we can find the coefficients \(c_n\) of the power series.

Step 4 ：The power series expansion of \(e^{-7x^2}\) is given by: \[1 - 7x^2 + \frac{49x^4}{2} - \frac{343x^6}{6} + O(x^7)\] This means that the coefficients \(c_n\) of the power series solution of the differential equation are given by: \[c_0 = 1, \quad c_1 = 0, \quad c_2 = -7, \quad c_3 = 0, \quad c_4 = \frac{49}{2}, \quad c_5 = 0, \quad c_6 = -\frac{343}{6}\]

Step 5 ：The radius of convergence of a power series is the distance from the center of the series to the nearest singularity of the function. In this case, the function \(e^{-7x^2}\) has no singularities, so the radius of convergence is infinity.

Step 6 ：Final Answer: The power series solution of the differential equation is given by: \[y(x) = 1 - 7x^2 + \frac{49x^4}{2} - \frac{343x^6}{6} + \cdots\] The coefficients of the power series are \(c_0 = 1\), \(c_1 = 0\), \(c_2 = -7\), \(c_3 = 0\), \(c_4 = \frac{49}{2}\), \(c_5 = 0\), and \(c_6 = -\frac{343}{6}\). The radius of convergence of the series is infinity. Therefore, the power series represents the function \(e^{-7x^2}\) for all real values of \(x\). \(\boxed{y(x) = 1 - 7x^2 + \frac{49x^4}{2} - \frac{343x^6}{6} + \cdots}\)