Trying to solve a second order ODE

cahill8
Messages
29
Reaction score
0

Homework Statement



I'm trying to solve a second order ODE for y(x) to show that the solution is y(x)=sin(x)/x. We've been told to use the substitution y(x)=h(x)/x. I've got to the stage of solving for h(x), arriving at h''(x)=-x. Using the general solution, h(x)=A sin(x) + B cos(x) and substiting this into the original equation for y(x) I get y(x)=(A sin(x) + B cos(x))/x

So all that's left to do it seems is to use the inital conditions to show A=1 and B=0 however the problem says to use the inital condition y(0)=1. This doesn't make sense to me since y is singular at x=0. Is this inital condition a mistake or am I missing something?


Homework Equations


given above


The Attempt at a Solution


given above
 
Physics news on Phys.org
hi cahill8! :smile:

limx->0 sinx/x = 1 :wink:
 
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