For what values of k does y=5+3e^(kx) satisfy the differential equation dy/dx= 10-2y

I get k=1 but the book says -2...

My steps were: find y' and plug it into the left side of the equation and then plug y into the right side. This gave me 3e^(kx) = 3ke^(kx)... Where did I go wrong?

Question

For what values of k does y=5+3e^(kx) satisfy the differential equation dy/dx= 10-2y

I get k=1 but the book says -2...

My steps were: find y' and plug it into the left side of the equation and then plug y into the right side. This gave me 3e^(kx) = 3ke^(kx)... Where did I go wrong?

anonymous · Answer

But using your method:
Since $y=5+3e^{kx}$, you have $y'=3ke^{kx}$. Subbing into the equation, you have
$$3ke^{kx}=10-2(5+3e^{kx})\
3ke^{kx}=-6e^{kx}\
3k=-6$$

anonymous · Answer

I haven't formally learned diff eqs yet, taking it in the fall so I'm just doing the intro stuff in my calc book. And thanks, that's twice I have failed at distributing a simple constant! Gah!

anonymous · Answer

You're welcome!