how to solve this:

An approximate solution of the problem $y^"-y^{'}+4x\epsilon^x =0$, $y^{'}(0)-y(0)=1$,$y^{'}(1)+y(1)=-\epsilon$ is: here we have to calculate the value of y(x)?

what i did is: for this we use rayleigh-Ritz method, and i got the function $f(x,y,y^{'},y^{"})=y^{'2}+y^2-8x\epsilon ^x y^{"}$

i assumed the function y(x)=$c_1+c_2x$

how to proceed further and the answer and am not able to get the correct value of the constants?

  • $\begingroup$ I probably miss something but why don't you solve for $y$ ? It is a first order differential equation and the solution is simple. $\endgroup$ – Claude Leibovici Sep 1 '14 at 10:02
  • $\begingroup$ how to solve it? $\endgroup$ – amit Sep 1 '14 at 11:43
  • $\begingroup$ Start defining $z=y'$. You have a first order differential equation in $z$; solve and integrate again. $\endgroup$ – Claude Leibovici Sep 1 '14 at 16:02
  • $\begingroup$ This does not look like a typical problem for Rayleigh-Ritz. Are you sure you have (1) the equation stated correctly, (2) the boundary conditions stated correctly? You say the answer is 0.58+0.27x. That is hopeless for the boundary conditions you have given. $\endgroup$ – almagest Sep 6 '14 at 9:46

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