Find the power series solution of the form y=a0+a1x+a2x^2++anx^n+ to the differential equations below
a) xy''+4y=0
b) y"'+y'+y=0
give the first six non zero terms in each solution.
you

b) is y''+y'+y=0

For both ODEs, we take

which has derivatives

Notice that we treat in such a way as to have , , and

a.

Substituting the relevant series above into the ODE gives

which immediately tells us , or , and for every other coefficient we get

So we have

and so on, with the general pattern

Then the first 6 non-zero terms of the solution are

# # #

b. I'm not sure if you mean to include the third derivative, or if the last part of your post is correcting it to be the second derivative. I'll assume the latter because it's slightly simpler.

Substituting the series above into the ODE gives

so that

Finding a general pattern here is a bit more involved, so we'll compute the coefficients up to 6 non-zero terms:

which are all non-zero as long as and , so the first 6 non-zero terms are

# # #

Included below are plots of the exact solutions to both ODE (in blue) along with the first 6 terms found above, with assumed initial conditions.

In a relation h(x) = y, the values of x are the domain and the values of y are the range.

Here the domain is:

D = {-6, -1, 7}

The range is:

R = {-9, 0 , 9}

Now, a relation f(x) = y is a function only if for every x in the domain we have only and only one value in the range.

Here we can see that for the value -1 in the domain we have two different values in the range:

(-1, 0) and (-1, 9)

So this can not be a function.

(If you want to take the variable as y, you also have that the value y = 0 leads to two different values in x, so this cant be a function either)