# Taylor Series Solutions of Differential Equations

Hi there, I have a question regarding Taylor Series ODE which can be located here http://nullspace8.blogspot.com/2011/10/14.html. can someone please take a look? Full working step by step solution in pdf or word please. If you think the bid is insufficient and you can do it, please respond with a counter bid. Thank you.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

a) We have y' = x^2 y^2 with initial condition y(0) = 1 and we wish to compute the Taylor series of y(x) up to n=2. First we write

y = c_0 + c_1 x + c_2 x^2 + O(x^3).

Since y(0) = 1 we have c_0 = 1. Now

y' = c_1 + 2 c_2 x + O(x^2).

But this is also equal to

x^2 y^2

= x^2(c_0 + c_1 x + c_2 x^2 + O(x^3)

= c_0 x^2 + c_1 x^3 + c_2 x^4 + O(x^5).

Equating terms multiplying 1 and x, we see that c_1 = c_2 = 0, whence

y = 1 + O(x^3).

(b) We have y' = x^2 + y^2 with initial condition y(0) = 1 and we wish to compute the Taylor series of y(x) up to n=3. We write

y(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + O(x^4).

Since y(0) = 1, we have c_0 = 1. We also have

y' = c_1 + 2 c_2 x + 3 c_3 x^2 + O(x^3).

But this is also equal to

x^2 + y^2

= x^2 + (c_0 + c_1 x + c_2 x^2 + O(x^3))^2

= x^2 + c_0^2 + 2 c_0 c_1 x + (2 c_0 c_2 + c_1^2) x^2 + O(x^3)

= c_0^2 + 2 c_0 c_1 x + (1 + 2 c_0 c_2 + c_1^2) x^2 + O(x^3).

Equating terms multiplying powers of x up to x^2, we find

c_1 = c_0^2 = 1

2 c_2 = 2 c_0 c_1 = 2 c_1 = 2, whence c_2 = 1

3 c_3 = 1 + 2 c_0 c_2 + c_1^2

= 1 + 2 + 1 = 4, whence c_3 = 4/3.

Thus we have

y(x) = 1 + x + x^2 + 4/3 x^3 + O(x^4).

(c) We have

y' = sin(xy), where y(0) = 1

and we wish to compute the Taylor series of y up to n=2.

We have

y(x) = c_0 + c_1 x + c_2 x^2 + O(x^3).

From the initial condition we have c_0 = 1. We also have

y' = c_1 + 2 c_2 x + O(x^2)

= sin(xy) = xy + O(x^3)

= c_0 x + O(x^2).

Thus we have c_1 = 0 and 2 c_2 = c_0 = 1, whence c_2 = 1/2, so

y(x) = 1 + 1/2 x^2 + O(x^3).

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