Question

The solution to the initial value problem

$\begin{array}{l}x{y}^{\text{'}\text{'}}\left(x\right)+2y^{\text{'}}\left(x\right)+xy\left(x\right)=0\\ y\left(0\right)=1,y^{\text{'}}\left(0\right)=0\end{array}$

has derivatives of all orders at$\mathit{x}\mathbf{=}\mathbf{0}$(although this is far from obvious). Use L'Hôpital's rule to compute the Taylor polynomial of degree 2 approximating this solution.

Short answer

The required polynomial is,$p_3\left(x\right)=1-\frac{x^2}{6}$ .

Step-by-step solution

Step 1:To Find the Taylor polynomial of degree

The formula for the Taylor polynomial of degree n centered at ,$x_0$approximating a function$f\left(x\right)$possessing n derivatives at , $x_0$is given by

$p_n\left(x\right)=f\left(x_0\right)+f^{\text{'}}\left(x_0\right)\times (x-x_0)+f^{\text{'}\text{'}}\left(x_0\right)\times \frac{{(x-x_0)}^2}{2!}+\cdots +f^n\left(x_0\right)\times \frac{{(x-x_0)}^n}{n!}$

The differential equation is given as

$x{y}^{\text{'}\text{'}}\left(x\right)+2y^{\text{'}}\left(x\right)+xy\left(x\right)=0$

It is given that for the function,$y\left(x\right)$

$y\left(0\right)=1\text{and}{y}^{\text{'}}\left(0\right)=0$

For applying the L’Hospital rule we need to see that our differential equation satisfies $\frac{0}{0}$form,

$\begin{array}{c}x{y}^{\text{'}\text{'}}\left(x\right)+2y^{\text{'}}\left(x\right)+xy\left(x\right)=0\\ y^{\text{'}\text{'}}\left(x\right)+\frac{2y^{\text{'}}\left(x\right)}{x}+y\left(x\right)=0\\ y^{\text{'}\text{'}}\left(x\right)=-y\left(x\right)-\frac{2y^{\text{'}}\left(x\right)}{x}\\ y^{\text{'}\text{'}}\left(x\right)=\frac{-xy\left(x\right)-2y^{\text{'}}\left(x\right)}{x}\\ y^{\text{'}\text{'}}\left(0\right)=0\end{array}$

Hence, it satisfies L'Hospital rule and by applying it in differential equation it becomes,

$\begin{array}{c}{y}^{\text{'}\text{'}}\left(x\right)=\frac{-xy\left(x\right)-2y^{\text{'}}\left(x\right)}{x}\\ y^{\text{'}\text{'}}\left(x\right)=-2y^{\text{'}\text{'}}\left(x\right)-y\left(x\right)-x{y}^{\text{'}}\left(x\right)\\ y^{\text{'}\text{'}}\left(0\right)=-2y^{\text{'}\text{'}}\left(0\right)-y\left(0\right)-x{y}^{\text{'}}\left(0\right)\\ y^{\text{'}\text{'}}\left(0\right)=\frac{-1}{3}\end{array}$

Step 2:Final proof

The Taylor polynomial of degree 2 in the solution is given by.$p_3\left(x\right)=1-\frac{x^2}{6}$