Using "bvp4c" to solve 2nd order DE?

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Aaron
Aaron on 11 Sep 2013
I was told that using a template and changing a few things would provide me with an answer to a 2nd order DE, using bvp4c. But I'm not sure how. I changed my low and high boundaries
function bvp4c
xlow = 0; xhigh = 10;
solinit = bvpinit(linspace(xlow,xhigh,10),[1 -1]);
sol = bvp4c(bvp4ode,bvp4bc,solinit);
xint = linspace(xlow,xhihg,20);
sxint = deval(sol,xint);
plot(xint,sxint(1,:))
function dydx = bvp4ode(x,y)
dydx = [y(2) y(1)];
function res = bvp4bc(ya,yb)
res = [ya(1) yb(1)]; % I think this is another area of concern.
Here is the initial data given: d^2T/dx^2 - 0.15T = 0 T(0) = 240 T(10) = 150
This is a bit confusing!!

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