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The answer I got for this was y =5/2 + (y_0-5/2)e^t

but the solution is y =5/2 + (y_0-5/2)e^-2t

Where did this -2t come from?

My steps:

dy = (2y-5)dt Integrating both sides gives

1/2*ln|2y - 5| = t + C take exp of both sides gives

1/2(2y-5) = Ce

^{t}

y-5/2 = Ce

^{t}

solving for C

(y

_{0}) - 5/2=C*e

^{0}

C= y

_{0}- 5/2