Finding Intersection Points between 3rd Order ODE and a line

Question

Izu 2011-2-7

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https://ww2.mathworks.cn/matlabcentral/answers/986-finding-intersection-points-between-3rd-order-ode-and-a-line

How can I find the intersection points between the solution of a 3rd order ODE and a line y=x?

My ODE's code is

sol=dsolve('D3y-4*D2y+Dy+2*y=0,y(0)=-4,Dy(0)=-6,D2y(0)=-4')
x=0:2
y=subs(sol,'t',x)
plot(x,y)

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Answer 1

Andrew Newell 2011-2-7

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https://ww2.mathworks.cn/matlabcentral/answers/986-finding-intersection-points-between-3rd-order-ode-and-a-line#answer_1406

在 MATLAB Online 中打开

The solution of this equation is a symbolic function sol(t):

sol=dsolve('D3y-4*D2y+Dy+2*y=0,y(0)=-4,Dy(0)=-6,D2y(0)=-4');

sol(t)=t is the same as sol(t)-t = 0:

syms t
functionToBeZeroed = sol - t;

Turn this into a numerical function f(x):

f = matlabFunction(functionToBeZeroed);

You can calculate f(x) for any value of x. A plot shows that the curve crosses zero near about x=1.6.

x = 0:.01:2;
plot(x,f(x))

So use fzero to find the solution of sol(x)=x with an initial guess of x=1.6:

fzero(f,1.6)

EDIT: This answer has been revised to expand the explanatory text.

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Izu 2011-2-8

Uh... sorry... I don't get it... Let's try for a simple function: t = 0:.3:10;

y = sin(t); How will we find the intersections with y=x?

Andrew Newell 2011-2-8

I have added some more explanation above. Please note that your simple function is actually sin; the argument t is just the name you choose for the input and the specific values t=0:.3:10 are just a set of numbers you could calculate the sine of. It would be the same to calculate y=sin(x) for x=0:.3:10. To solve sin(x)=x, define f(x) = sin(x)-x and find the zero.

If, on the other hand, you are really trying to solve sin(t)=x, the question is meaningless.

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