Question

I'm trying to solve some differential equations, and I think I have the right answer, but I'm not sure :( With two equations.

Answer 1

These are the two equations I'm dealing with. I think I have the first one so far.

Answer 2

what type
quadratic, simulatneous?

Answer 3

\[\frac{ dC }{ dt }=-k _{1}C \]

Answer 4

\[\int\limits_{}^{}\frac{ dC }{ C } = \frac{ -k _{1} Cdt}{ C}\]

Answer 5

\[\int\limits_{}^{}\frac{ dC }{ C }=\int\limits_{}^{}-k _{1}dt\]

Answer 6

\[\ln (C)=-k _{1}\int\limits_{}^{}dt\]

Answer 7

\[\ln (C)=-k _{1}t+C _{1}(constant of integration)\]

Answer 8

I'm not sure how to continue from here, or if this is right :|

Answer 9

e^(lnx) = x

Answer 10

\[\ln (C)=-k _{1}t+\delta\]
\[exp(\ln (C))=exp(-k _{1}t+\delta)\]
\[C=exp(-k _{1}t+\delta)\]
\[C=exp(-k _{1}t)~exp(\delta)\]
\[C=\lambda~exp(-k _{1}t)\]

Answer 11

when t=0, C = lambda = Ao to satisfy your initial condition therefore:
\[\large C=A_oe^{-k_1t}\]

Answer 12

\[\large \frac{dG}{dt}=k_1~A_oe^{-k_1t}-k_2G+k_2G_o\]
that last term is a constant
\[\large G'+k_2G=k_1~A_oe^{-k_1t}+k_2G_o\]

Answer 13

any of this make sense?

Answer 14

Hey! Thank you so much for the help, I'm sorry I didn't get back to you sooner, I was in class. But yes, that makes sense to me! We didn't really go over much of this in class so I was having trouble getting it. Thanks again!!