Question

Find power series solutions in powers of \(x\) of each of the differential equations in Exercises. $$ \left(x^{2}+1\right) y^{\prime \prime}+x y^{\prime}+x y=0 $$

Short answer

The power series solution for the given ODE \(\left(x^{2}+1\right) y^{\prime \prime}+x y^{\prime}+x y=0\) can be found by assuming the solution in the form of a power series \(y(x) = \sum_{n=0}^{\infty} a_n x^n\) and calculating its first and second derivatives. Substitute the power series and their derivatives into the given ODE and combine the sums. The coefficients of the power series can be determined iteratively using the relation \(a_{n+2} = -\frac{(n+1)a_{n+1} + a_{n+1}}{n(n-1)}\). The final solution will have the form \(y(x) = a_0 y_1(x) + a_1 y_2(x)\), where \(y_1(x)\) and \(y_2(x)\) are linearly independent power series solutions constructed using the coefficients found from the relation provided.

Step-by-step solution

Assume a power series solution for y(x)

Assume the solution of the given ODE can be expressed as a power series: \[y(x) = \sum_{n=0}^{\infty} a_n x^n\] where \(a_n\) are the coefficients of the power series.

Calculate the derivatives

Calculate the first and second derivatives of the assumed power series solution: First derivative: \[ y'(x) = \sum_{n=1}^{\infty} na_n x^{n-1} \] Second derivative: \[ y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} \]

Substitute the power series and derivatives into the given ODE

Substitute the power series solutions and their derivatives into the given ODE: \[ \left(x^{2}+1\right) \left( \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} \right)+x \left( \sum_{n=1}^{\infty} na_n x^{n-1} \right)+x \left( \sum_{n=0}^{\infty} a_n x^n \right)=0 \]

Combine and rearrange the sums

Combine the sums, and shift indices if necessary, to make it easier to find the relationship between the coefficients: \[ \sum_{n=0}^{\infty} [n(n-1)a_{n+2} + (n+1)a_{n+1} + a_{n+1}]x^n = 0 \]

Solve for the coefficients

Since the equality holds for all values of \(x\), the coefficients must satisfy the following relation: \[ n(n-1)a_{n+2} + (n+1)a_{n+1} + a_{n+1} = 0 \] Now, for each \(n\geq0\), we can find an expression for \(a_{n+2}\): \[ a_{n+2} = -\frac{(n+1)a_{n+1} + a_{n+1}}{n(n-1)} \] The coefficients of the power series solution can be determined iteratively, starting from \(a_0\) and \(a_1\), which will act as arbitrary constants (since it is a second-order ODE, there will be two arbitrary constants). The final solution will then have the form: \[ y(x) = a_0 y_1(x) + a_1 y_2(x), \] where \(y_1(x)\) and \(y_2(x)\) are linearly independent power series solutions constructed using the coefficients we have just found.