2 years, 3 months ago
Simultaneous equation
Help!, I have:
y=e^-2x (C*cos(2x)+D*sin(2x)) - cos(2x) - sin(2x)
(thats e to the power of -2x, and then multiplied by the contents of the brackets)
and I'm given that when x = 0 then y = 1.
The second equation is the derivative of the first. i.e.
dy/dx = e^-2x (2D cos(2x) - 2C sin(2x)) - 2e^-2x (C cos(2x) + D sin(2)) + 2 sin(2x) - 2 cos(2x)
and for this I'm give that when x = 0 then dy/dx = -6
So with the two equations you can solve for C and D - that's the question! I am not sure of the result I'm getting for D (I think its either -1 or 0 or it could be any value at all!, putting x = 0 in D*sin(2x) results in D be eliminated from the first equation, does that mean it can take any value?)
y=e^-2x (C*cos(2x)+D*sin(2x)) - cos(2x) - sin(2x)
(thats e to the power of -2x, and then multiplied by the contents of the brackets)
and I'm given that when x = 0 then y = 1.
The second equation is the derivative of the first. i.e.
dy/dx = e^-2x (2D cos(2x) - 2C sin(2x)) - 2e^-2x (C cos(2x) + D sin(2)) + 2 sin(2x) - 2 cos(2x)
and for this I'm give that when x = 0 then dy/dx = -6
So with the two equations you can solve for C and D - that's the question! I am not sure of the result I'm getting for D (I think its either -1 or 0 or it could be any value at all!, putting x = 0 in D*sin(2x) results in D be eliminated from the first equation, does that mean it can take any value?)
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M$1 Answer
Why not simply substitute x=0 into the first equation which gets you:
y=e^0 (C*cos(0)+D*sin(0)) - cos(0) - sin(0) = C - 1 = 1
from which C = 2
Then, substitute C = 2 and x =0 into the second equation, getting:
dy/dx = e^0 (2D cos(0) - 4 sin(0)) - 2e^0 (2 cos(0) + D sin(0)) + 2 sin(0) - 2 cos(0) = 2D - 4 + 2 = -6
from which 2D = -4 or D = -2
The fact that D disappears for one value of x in the first equation just means that at that particular point (x=0) D can be anything. However, your second equation constrains D as you can see from the above.
y=e^0 (C*cos(0)+D*sin(0)) - cos(0) - sin(0) = C - 1 = 1
from which C = 2
Then, substitute C = 2 and x =0 into the second equation, getting:
dy/dx = e^0 (2D cos(0) - 4 sin(0)) - 2e^0 (2 cos(0) + D sin(0)) + 2 sin(0) - 2 cos(0) = 2D - 4 + 2 = -6
from which 2D = -4 or D = -2
The fact that D disappears for one value of x in the first equation just means that at that particular point (x=0) D can be anything. However, your second equation constrains D as you can see from the above.
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M$
I see my mistake now. My apologies. Here is the corrected version of the second equation:
dy/dx = e^0 (2D cos(0) - 2*2 sin(0)) - 2e^0 (2 cos(0) + D sin(0)) + 2 sin(0) - 2 cos(0) = 2D - 4 - 2 = -6
from which D=0.
I think it's 2D - 4 - 2 =-6
so D=0
Not really
2sin(0)-2cos(0) is separated of - 2e^0 (2 cos(0) + D sin(0))
Everything you did is correct except that -2cos(0) is -2, not +2
Note that there is a -2 multiplying the entire 4 term expression
(2 cos(0) + D sin(0)) + 2 sin(0) - 2 cos(0))
(there was a close-parentheses missing)