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dy/dx=2y+6, y(1)=6 (calculus II) the answer is y(x)=9e^(2x-2)-3

Mathematics
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explain what u need to find
\[\large \frac{dy}{dx}=2y+6\] So this appears to be separable :) Divide each side by (2y+6), and move the dx to the other side, think of this step as multiplication. \[\large \frac{dy}{2y+6}=dx\] Understand that part? From here we can integrate both sides.
yes, got that part!

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Other answers:

\[\large \int\limits \frac{dy}{2y+6}=\int\limits dx\]Integrating the right side is pretty straight forward,\[\large \int\limits\limits \frac{dy}{2y+6}=x+c\] Do you understand how to do the left side? You can apply a `u sub` if it will help.
go ahead ur doing great @zepdrix
all I need is the steps, i understand most of it.
Just in case you're having trouble with the left side, a u substitution would look like this, \[\large u=2y+6 \qquad \qquad \qquad \frac{1}{2}du=dy\] Which changes our integral to,\[\large \frac{1}{2}\int\limits\frac{du}{u}=x+c\] Giving us,\[\large \frac{1}{2}\ln|2y+6|=x+c\]
I already got that far. Now to put in the form y=9e^(2x-2)-3.
Multiply both sides by 2,\[\large \ln|2y+6|=2x+c\] We distributed the 2 to each term, but we simply `absorbed` the other 2 into the C, since it's an arbitrary value. Now exponentiate both sides, `rewrte both sides as exponents with a base of e`.\[\huge e^{\ln|2y+6|}=e^{2x+c}\]
The exponential and the logarithm are `inverse` operations of one another, so on the left they will simply "undo" one another.\[\huge 2y+6=e^{2x+c}\]
oh, i get it!
Understand how to solve for C from here? :)
i forgot that c was a constant.
and could be treated like that. now it's easy i think.
but what about the 9?
Are you not coming up with a 9? :o
y=e^(2x+c)/2-3
There's what I've got so far.
Ok cool. You need to solve for C at some point :) That's where your 9 will come from.
Don't fully comprehend, could you please help a little more.
At the start they gave us a coordinate pair that we can NOW use to solve for C.\[\large y(1)=6 \qquad \qquad (1,6)\] Plug the 1 in for X, the 6 in for Y, and solve for C! :) I have a feeling this will be a tad easier to work with if we fiddle with the C first. Using a rule of exponents we can write the C like this. \[\huge 2y+6=e^{2x}e^c\] But now recognize, that e^c is just another arbitrary constant value.\[\huge 2y+6=Ce^{2x}\]Then subtract, divide,...\[\huge y=Ce^{2x}-3\]Again the 2 got absorbed into the C. This isn't totally necessary. I just think the problem will be a lot easier to work with from this form.
\[\huge (\color{royalblue}{1},\color{orangered}{6}) \qquad \rightarrow \qquad \color{orangered}{y}=Ce^{2\color{royalblue}{x}}-3\]Understand how to plug these values in? :)
I finally got it!!! Thanks a lot!
yay! \c:/
try solving mine plz
The silly bread problem? c:
ya

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