Question

Find the solution to the differential equation y'=-y if the initial condition is y=3 and x=4.

Answer 1

\[\frac{dy}{dx}=-y \implies \frac{dy}{y}=-dx \implies \int\limits_3^y \frac{dy'}{y'}=- \int\limits_4^x dx'=\]
\[\ln(\frac{y}{3})=-(x-4) \implies y=3 e^{-(x-4)}\]

Answer 2

Or:
\[y'+y=0 \implies \lambda+1=0 \implies \lambda=-1 \implies \]\[y(x)=Ae^{-x} \implies 3=Ae^{-4} \implies A=3e^4 \implies y(x)=3e^4e^{-x}=3 e^{-(x-4)}\]

Answer 3

thank you so much for the step by step solution!!

Answer 4

No problem :) I like staying tip top on my diffeq/multivar/vector calc/etc.