Hi,
I've got the following ODE:
y''-ay'=b
that should satisfy:
y''(0)=e
y'(0)=0
I get the following for y''-ay'=0 :
y(x)=c1+c2*exp(ax)
For the nonhomogeneous part I use "undetermined coefficients method":
I guess Yp=c (a polynom of order 0) and when substituting into the diff eq I get: 0 - 0 = b.
Question 1: What does that mean? Is the solution: y(x)=c1+c2exp(ax) (that is, no contribution comming from b?)
Now say I wish to satisfy the boundary conditions:
y'=a*c2*exp(ax)
y''=(a^2)*c2*exp(ax)
for y''(0)=e:
(a^2)*c2=e --> c2=e/(a^2)
for y'(0)=0:
a*c2=0 --> c2=0
so...
Question 2: How do I find c1 and c2?
Thanks!
All the best
Roy
I've got the following ODE:
y''-ay'=b
that should satisfy:
y''(0)=e
y'(0)=0
I get the following for y''-ay'=0 :
y(x)=c1+c2*exp(ax)
For the nonhomogeneous part I use "undetermined coefficients method":
I guess Yp=c (a polynom of order 0) and when substituting into the diff eq I get: 0 - 0 = b.
Question 1: What does that mean? Is the solution: y(x)=c1+c2exp(ax) (that is, no contribution comming from b?)
Now say I wish to satisfy the boundary conditions:
y'=a*c2*exp(ax)
y''=(a^2)*c2*exp(ax)
for y''(0)=e:
(a^2)*c2=e --> c2=e/(a^2)
for y'(0)=0:
a*c2=0 --> c2=0
so...
Question 2: How do I find c1 and c2?
Thanks!
All the best
Roy