The function y satisfies a differential equation of the form y' = ky for some number k. If you are told that when t = 3 that y is 2 and the rate of change of y is 4 then what is k?

Accepted Solution

[tex]y'=ky[/tex] is a separable ODE:[tex]\dfrac{\mathrm dy}{\mathrm dt}=ky\implies\dfrac{\mathrm dy}y=k\,\mathrm dt[/tex]Integrating both sides and solving for [tex]y[/tex], we get[tex]\ln|y|=kt+C\implies y=e^{kt+C}\implies\boxed{y=Ce^{kt}}[/tex]Given that [tex]y(3)=2[/tex], we know[tex]2=Ce^{3k}[/tex]We also have[tex]y'=Cke^{kt}[/tex]and given that [tex]y'(3)=4[/tex], we know[tex]4=Cke^{3k}[/tex]Then[tex]Cke^{3k}=2Ce^{3k}\implies Ck=2C\implies k=2\text{ or }C=0[/tex]If [tex]k=2[/tex], then [tex]2=Ce^6\implies C=2e^{-6}[/tex].If [tex]C=0[/tex], then we get a contradiction because we need to have [tex]Ce^{3k}=2[/tex].So it must the case that [tex]k=2[/tex].