Differential Equations - Initial Value Problem

mattbonner
Messages
14
Reaction score
0

Homework Statement



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) =

Homework Equations



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

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
 
Physics news on Phys.org
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.
 
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:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top