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Mathematics
Differential Equations
Using complex numbers or phasor transform to solve O.D.E's
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[QUOTE="Mark44, post: 6007064, member: 147785"] By the way, there are at least two other ways to solve the first DE you listed, both of which are simpler. Here's one of them. 4y' + 2y = 10cos(x) The related homogeneous DE is 4y' + 2y = 0, or equivalently 2y' + y = 0 This is a linear, constant coefficent homogeneous DE. Try a solution of the form ##y = e^{rx}## Then ##y' = re^{rx}## Substituting, we get ##2re^{rx} + e^{rx} = 0 \Rightarrow (2r + 1)e^{rx} = 0## Solve for r to get r = -1/2 Solution of homogeneous problem: ##y_h = Ce^{-x/2}## For the nonhomogeneous problem 4y' + 2y = 10cos(x), try a solution of the form ##y_p = A\cos(x) + B\sin(x)##. Substitute this into the DE and solve for the coefficients A and B. The general solution will be ##y_h + y_p##, or in this case, ##y = Ce^{0.5x} + 1\cos(x) + 2\sin(x)## To get a value for the constant C, you need an initial condition. [/QUOTE]
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Using complex numbers or phasor transform to solve O.D.E's
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