Simple ODE. Boundary conditions are messing with my head... Please help me

[Roy]

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

 

[HallsofIvy]

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)

Okay, that is correct.

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.

A constant is already a solution to the associated homogeneous equation. You should try, instead, multiplying by x:
Yp= cx. Yp'= c, Yp''= 0 so you get 0- ac= b and c= -b/a.

Question 1:

What does that mean? Is the solution: y(x)=c1+c2exp(ax) (that is, no contribution comming from b?)

No, it means that was not a good choice!

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

Using \(\displaystyle y(x)= C_1+ C_2e^{ax}- (b/a)\), you have \(\displaystyle y(0)= C_1+ C_2= 0\) and \(\displaystyle y''(0)= b- ay'(0)= b- aC_2= 0\).

Solve those equations for \(\displaystyle C_1\) and \(\displaystyle C_2\).

 

[Roy]

Thank-you-so-much!

This was great help
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Just to verify with you:

Y=c1+c2*exp(a*x)-(b/a)*x

Right?


And, my BC is y'(0)=0 and y''(0)=e. Is it ok that I have no BC on y?

Thanks,
Roy