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verify that y=e^-x is a solution to the differential equation 3y'+4y=e^-x
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do you know how to take the derivative (with respect to x) of e^-x ?
Yes, (-1/e^x)
@phi
I would leave it -e^-x (not that it matters too much) replace y' with -e^-x replace y with e^-x and check that the left side matches the right side in your equaiotn
yes, I did that, then what
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you are trying to see if both sides are the same when you plug in your y' and y with -e^(-x) and e^(-x) respectively
if both sides are the same then y=e^(-x) is a solution if both sides aren't the same then y=e^(-x) is not a solution
I get -3/e^x +4e^-x = e^-x
I got the answer, thank you everyone
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