Question

Find the power series in \(x\) for the general solution. $$ \left(1+2 x^{2}\right) y^{\prime \prime}+7 x y^{\prime}+2 y=0 $$

Short answer

The method used to solve the given differential equation is the method of power series. The power series for the general solution is: $$ y(x) = a_2\left(-4x + x^2 -\frac{6}{4}x^3 +\dots\right) $$

Step-by-step solution

Assume a power series solution

Assume a power series solution of the form: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$

Differentiate the power series solution

Differentiate the power series solution twice with respect to \(x\) to obtain \(y^{\prime}(x)\) and \(y^{\prime \prime}(x)\): $$ y^{\prime}(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} $$ and $$ y^{\prime \prime}(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} $$

Substitute into the differential equation

Substitute \(y\), \(y^{\prime}\), and \(y^{\prime \prime}\) into the given differential equation and simplify: $$ \left(1+2x^2\right)\left(\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\right) + 7x \left(\sum_{n=1}^{\infty} n a_n x^{n-1}\right) + 2\left(\sum_{n=0}^{\infty} a_n x^n\right) = 0 $$

Equate coefficients of like powers of x

Equate coefficients of like powers of \(x\) to form a recurrence relation for the coefficients \(a_n\). For example, for \(x^0\), \(x^1\), and \(x^2\), we obtain: $$ 2 a_0 = 0,\quad 1 a_1 + 4 a_2 = 0,\quad 2 a_0 + 3(2)a_1 + 6a_2 + 4a_3 = 0 $$ Solving these equations, we find: $$ a_0 = 0,\quad a_1 = -4a_2, \quad a_3 = -\frac{6a_2}{4} $$ Use these results to generate the power series solution to the given differential equation.

Write the general solution

From the previous step, we obtained the relationships between the coefficients \(a_n\). By plugging them into the power series solution, we get: $$ y(x) = a_2\left(-4x + x^2 -\frac{6}{4}x^3 +\dots\right) $$ This is the power series for the general solution to the given differential equation.