L'Hopitals rule to solve 2nd order DE

[morenogabr]

Homework Statement

The solution to the initial value problem
x y''(x) + 2y'(x) + xy(x) = 0; y(0)=1, y(0)=0
has derivatives of all orders at x=0 (although this is far from obvious). Use L'Hopitals rule to compute the taylor polynomial of degree 2 approximating this solution.

Homework Equations

The Attempt at a Solution

I attempted divide all terms by x so as to solve for y'' and begin plugging y, y', y'', etc into formula for taylor series. Got stuck at x=0 in the denominators, i guess is where the far from obvious comes in... WWL'HD?

 

[mighty2000]

Thank you for your post. In order to compute the Taylor polynomial of degree 2 for the solution to the given initial value problem, we can use L'Hopital's rule to simplify the expression at x=0. This will allow us to find the derivatives of all orders at that point and thus compute the Taylor polynomial.

First, let's divide both sides of the equation by x to get:

y''(x) + 2y'(x)/x + y(x) = 0

Now, let's take the limit as x approaches 0 of both sides. By L'Hopital's rule, we can take the derivative of the left side with respect to x and the right side will remain the same. This gives us:

lim x->0 y''(x) + 2y'(x)/x + y(x) = lim x->0 0

Next, we can apply L'Hopital's rule again to the left side, taking the derivative with respect to x once more. This gives us:

lim x->0 y'''(x) + 2y''(x)/x + y'(x)/x = lim x->0 0

We can continue applying L'Hopital's rule until we have taken the derivative of the left side with respect to x n times, where n is the degree of the Taylor polynomial we are looking for. At each step, the limit on the right side will remain 0.

Now, we can plug in the initial conditions given in the problem to find the values of the derivatives at x=0. We know that y(0)=1 and y'(0)=0, so we can substitute these values into the equation above to get:

y''(0) + 2y'(0)/0 + y(0) = 0

Simplifying, we get:

y''(0) = -1

Similarly, we can find the values for the higher order derivatives at x=0. For example, using L'Hopital's rule one more time, we can find that:

y'''(0) = 0

Now, we can use these values to construct the Taylor polynomial of degree 2, which is given by:

P2(x) = y(0) + y'(0)x + y''(0)x^2/2!

Substituting in the values we found above, we