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y'' + 4y = 8 y(0)=2, y'(0)=1 Solve the following differential equation using the Laplace transform method.
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I know how to take the laplace transform of the left side but im not sure what to do with the 8 on the right side.
Just do \(8\mathcal L[1]\)
Oh that is right.. So it is basically 8t?
I guess? Can't remember the transforms.
actually.. 1 = 1/s so therefore it is 8/s
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\(1=t^0\)
I'm looking at m formula sheet and it says 1 = 1/s
Thanks anyways wio!
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