Write and solve the differential equation that models the verbal statement. (Use k for the constant of proportionality.) The rate of change of y is proportional to y. When x = 0, y = 10, and when x = 2, y = 25. What is the value of y when x = 4?
dy/dt=ky
dy/dx = ky y = ?? Evaluate the solution at the specified value of the independent variable. y = ??
wait dy/dx=ky
\[\frac{dy}{y}=kdx\]
Integrate both sides: \[\int{\frac{dy}{y}=\int k*dx}\]
\[ln|y|=kx+C\]
I think it's going ot be 25 =10e^k(2)
x = 0, y = 10 ln|10|=C \[ln|y|=kx+ln10\]
\[e^{ln|y|}=e^{kx}*e^{ln10}\] \[y=10e^{kx}\]
x=2, y=25 25=10*e^(2k) 2.5=e^(2k) ln(2.5)=2k k=ln(2.5)/2
\[\large{y=10*e^{\frac{ln2.5}{2}x}}\]
And to figure out what's y when x=4: y=10*e^(2*ln(2.5)), just use a calculator Hopefully I didn't make any silly mistakes
nope :).
Yay
62.5 yay!
lol when u said nope, I thought the answer didn't wrong... then i saw the smiley and got relieved hahah
20 more to go!
OMG
jk :P
Btw I have my own hw lol, I wasn't planning on sticking around. Still have to submit this stupid excel project, write a physics lab, and 2 programs. I have no idea why I'm wasting time here.
Got a handful :P. These are the toughes! I'm good in Physics, and what kinds of programs? :P
C++
oh yeah you were doing the c++ stuffs.
still fighting your teacher about the negating spaces? :P
I know it, it's just tedious... and very boring, and no, I gave up on her.. she can't teach at all
sux, I learned on my own....
lol I learned with a diff professor, I was retaking the same course basically
Lots of bad professors out there... That's why I did it all on my own... You learn much more than the classroom can teach you...
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