# Differential Equations - Initial Value Problem

1. Sep 11, 2009

### mattbonner

1. The problem statement, all variables and given/known data

Suppose that the initial value problem
y' = 7(x^2) + (5y^2) − 6, y(0)=−2
has a solution in an interval about x=0.

Find y'(0) =
Find y''(0) =
Find y'''(0) =

2. Relevant equations

get it into standard form: dy/dt + p(t)y = g(t)
find integrating factor = e^($$\int$$p(t)dt + k

multiply everything by integrating factor, simplify left-hand-side and then integrate both sides

using initial condition, solve for C
solve for y

3. The attempt at a solution

i don't seem to be able to get it into standard form

i tried doing y' - 5y^2 = 7x^2 -6
which gave me an integrating factor of e^-5xy

i tried following the rest of the steps with that integrating factor but its not working

2. Sep 11, 2009

### Dick

You don't actually have to solve the equation to find the derivatives of y(x) at x=0. To find y'(0) just plug x=0, y=(-2) into the equation. To find the higher derivatives, just differentiate the equation with respect to x a couple of times.

3. Sep 11, 2009

### mattbonner

oh wow i feel like such a moron
thank you so much!!

edit: wait, for y''(0)

i differentiated it, and i got 14x?

edit(2): nvm i solved it

Last edited: Sep 12, 2009