How to Solve a Linear ODE BV Problem with Boundary Values?

[mistereko]

Homework Statement

I nearly have this problem solved

x²d²y/dx² + 3x*dy/dx + 5y = 8x

y(1) = 2, y(exp(pi/4)) = 2sinh(pi/4)

I've found the general solution, but I'm not sure how to get the answer with the boundary values

Homework Equations

The Attempt at a Solution

My general solution is y(t) = C₁ e^-t cos(2t) + c₂ e^-t sin(-2t) + 2t

 

[mistereko]

I understand how to use BV's it just I'm getting weird answers

 

[mistereko]

I understand how to use BV's it just I'm getting weird answers

C₁ - C₂ = 6
C₁ - C₂ = 2sinh(pi/4) - 8e^pi/4

 

[Dick]

Homework Statement

I nearly have this problem solved

x²d²y/dx² + 3x*dy/dx + 5y = 8x

y(1) = 2, y(exp(pi/4)) = 2sinh(pi/4)

I've found the general solution, but I'm not sure how to get the answer with the boundary values

Homework Equations

The Attempt at a Solution

My general solution is y(t) = C₁ e^-t cos(2t) + c₂ e^-t sin(-2t) + 2t

It looks like you did a change of dependent variable like x=e^t to get a solution like that, right? And I'm not sure I agree with the ' + 2t' part. The initial conditions they are giving you are initial conditions for x, not for t. What values of t correspond to x=1 and x=exp(pi/4)?

 

[mistereko]

It looks like you did a change of dependent variable like x=e^t to get a solution like that, right? And I'm not sure I agree with the ' + 2t' part. The initial conditions they are giving you are initial conditions for x, not for t. What values of t correspond to x=1 and x=exp(pi/4)?

Yes, I used e^t. I'll try and work out new boundary values. I think you're right about the 2t part. Thanks

 

[mistereko]

Yes, I used e^t. I'll try and work out new boundary values. I think you're right about the 2t part. Thanks

t = 0 and t = pi/4.

 

[Dick]

t = 0 and t = pi/4.

Sure, now if you correct the '+ 2t' part to the correct expression you actually should be able to work out nice values for C1 and C2. Remember what the definition of sinh is.

 

[mistereko]

My general solution is y(t) = C1 e-t cos(2t) + c2 e-t sin(-2t) + e^t, but I don't know how to chnage the boundary values from x to t. Having a slow moment.

 

[mistereko]

can anyone verify that the final solution is y(t) = (cos(pi/2) + 2 + exp(pi/2)) e-t sin(2t) + e-t cos(2t) + e^t please?

 

[Dick]

can anyone verify that the final solution is y(t) = (cos(pi/2) + 2 + exp(pi/2)) e-t sin(2t) + e-t cos(2t) + e^t please?

Show how you got that.

 

[mistereko]

2 = c1*e^0*sin(0) + c2*e^0*cos(0) + e^0

that works out nicely to
C2 = 1

For C1 it's a bit messier. I let 2Sinh(pi/4) = (e^(pi/2) -1)/ e^(pi/4)

Therefore

(e^(pi/2) -1)/ e^(pi/4) = C1*exp(-pi/4)*sin(pi/2) + exp(-pi/4)*cos(pi/2) + exp(-pi/4)
multiplied by exp(pi/4)

and found c1 to be Cos(pi/2) + 2 -exp(-pi/2)

 

[mistereko]

No need, I just realized a huge mistake I made and figure it out. Thanks anyway :)

 

[Dick]

(e^(pi/2) -1)/ e^(pi/4) = C1*exp(-pi/4)*sin(pi/2) + exp(-pi/4)*cos(pi/2) + exp(-pi/4)

I'm taking it on trust you've got the right general solution there. But if so, shouldn't the last term be exp(pi/4) instead of exp(-pi/4)?

 

[mistereko]

Your dead right, it's been a very long night.

 

[mistereko]

Final solution is e^(-pi/4)*e^(-t)sin(2t) +e^(-t)cos2t + e^t

 

[Dick]

Final solution is e^(-pi/4)*e^(-t)sin(2t) +e^(-t)cos2t + e^t

Apparently it has been a long night. I just get C1=1 and C2=(-1), though you have been switching C1 and C2 around. And it might be good to express your final answer in terms of x rather than t.

 

[mistereko]

Just looked at it with fresh eyes.

y(x) = 1/x * cos(2ln(x)) - 1/x * sin(2ln(x)) + x

 

[Dick]

Just looked at it with fresh eyes.

y(x) = 1/x * cos(2ln(x)) - 1/x * sin(2ln(x)) + x

Looks ok. You can always try substituting into the original ODE if you've got any doubts.

 

[mistereko]

Brilliant, thanks again.