- #1
cameuth
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THE PROBLEM :
y(t) = e^(-t)*sin(t^2);
with t0 = 0 and T = 3.14159. Find y_0, and use it to deduce the corresponding expression
for f(t, y) (Your f should have both a t and a y in it. Simplify it to find the y!).
This is for a MATLAB project. I've solved this differential equation (which we get from the derivative of the problem above) , but apparently there's a way to solve it where we get y on the right hand side of the equation (and still have t's). This is the only way the code will work in matlab.
MY ATTEMPT: basically I took the derivative of y=e^(-t)*sin(t^2) to get
y'=-e^(-t)[sin(t^2)+(2t)cos(t^2)]
You then inegrate to get the original formula plus a constant
y=e^(-t)*sin(t^2)+C
Now if we apply the inital condition y(0)=0, we get the original equation. Again.
y=e^(-t)*sin(t^2)
This is not correct. What am I missing?
PS, if you want to see my code and think that will help I have that as well. Just ask. Thanks
y(t) = e^(-t)*sin(t^2);
with t0 = 0 and T = 3.14159. Find y_0, and use it to deduce the corresponding expression
for f(t, y) (Your f should have both a t and a y in it. Simplify it to find the y!).
This is for a MATLAB project. I've solved this differential equation (which we get from the derivative of the problem above) , but apparently there's a way to solve it where we get y on the right hand side of the equation (and still have t's). This is the only way the code will work in matlab.
MY ATTEMPT: basically I took the derivative of y=e^(-t)*sin(t^2) to get
y'=-e^(-t)[sin(t^2)+(2t)cos(t^2)]
You then inegrate to get the original formula plus a constant
y=e^(-t)*sin(t^2)+C
Now if we apply the inital condition y(0)=0, we get the original equation. Again.
y=e^(-t)*sin(t^2)
This is not correct. What am I missing?
PS, if you want to see my code and think that will help I have that as well. Just ask. Thanks
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