- #1
AlexCdeP
- 39
- 1
Hi guys, so I'm stuck on quite an interesting problem, and have been for a few days now. If anybody can take the time to have a look at it that would be the most incredible thing ever, because I have reached a point where I am at a loss.
Solve the following 4th order differential equation
20x^4y''''(x) - x^4y'''(x) + 3x^2y''(x) - 6xy'(x)+ 6y(x) =0
subject to the boundary conditions: y'(1)=0, y'''(1) = 1, y''(5) = -50,
and y'''(5)= -20. The superscripts within parentheses indicate the order
of the derivatives. Provide a plot of your solution y(x) from x = 1 to
x = 5.
I won't post my MATLAB code right now because it's very long, and I doubt any of you will want to wade through it! If anybody wants me to post it just say so. Instead I will write down what I think the algorithm is and I want you to call me out if you think I've made a mistake.
1. Rearrange the ode to solve for y'''' and then convert each differential into first order ODE. We have
function F = dEqs(y,x) % First-order differential
F=zeros(4,1);
F(1)=y(2);
F(2)=y(3);
F(3)=y(4);
F(4)= (1/20)*y(4)-(3/(20*x^2))*y(3)+(3/(10*x^3))*y(2)-(3/(10*x^4))*y(1);
2. Now this can at least be solved by ode45, BUT I don't have initial conditions, so I must guess my initial conditions. I need y''(1) and y(1), so I guess these (how I make a good guess I don't really know). This is called the shooting method for those who are interested. I can't use bvp4c because I need 3 out of 4 initial conditions I think. If I had 3 out of 4 I could solve it but...
3. THIS IS WHERE I AM STUCK I start with guessing y''(1) but I can't check this corresponds to y'(5) when I integrate because I don't know it. I can't guess y(1) for the same reason, in fact I don't know how to deal with that at all.
Any idea how I should go about solving this problem? I have a code for solving a second order ode, third or fourth with more forgiving boundary conditions... anybody?
Thank you!
Solve the following 4th order differential equation
20x^4y''''(x) - x^4y'''(x) + 3x^2y''(x) - 6xy'(x)+ 6y(x) =0
subject to the boundary conditions: y'(1)=0, y'''(1) = 1, y''(5) = -50,
and y'''(5)= -20. The superscripts within parentheses indicate the order
of the derivatives. Provide a plot of your solution y(x) from x = 1 to
x = 5.
I won't post my MATLAB code right now because it's very long, and I doubt any of you will want to wade through it! If anybody wants me to post it just say so. Instead I will write down what I think the algorithm is and I want you to call me out if you think I've made a mistake.
1. Rearrange the ode to solve for y'''' and then convert each differential into first order ODE. We have
function F = dEqs(y,x) % First-order differential
F=zeros(4,1);
F(1)=y(2);
F(2)=y(3);
F(3)=y(4);
F(4)= (1/20)*y(4)-(3/(20*x^2))*y(3)+(3/(10*x^3))*y(2)-(3/(10*x^4))*y(1);
2. Now this can at least be solved by ode45, BUT I don't have initial conditions, so I must guess my initial conditions. I need y''(1) and y(1), so I guess these (how I make a good guess I don't really know). This is called the shooting method for those who are interested. I can't use bvp4c because I need 3 out of 4 initial conditions I think. If I had 3 out of 4 I could solve it but...
3. THIS IS WHERE I AM STUCK I start with guessing y''(1) but I can't check this corresponds to y'(5) when I integrate because I don't know it. I can't guess y(1) for the same reason, in fact I don't know how to deal with that at all.
Any idea how I should go about solving this problem? I have a code for solving a second order ode, third or fourth with more forgiving boundary conditions... anybody?
Thank you!