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rosho
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dy/dx=2y+6, y(1)=6 (calculus II)
the answer is y(x)=9e^(2x2)3
 one year ago
 one year ago
rosho Group Title
dy/dx=2y+6, y(1)=6 (calculus II) the answer is y(x)=9e^(2x2)3
 one year ago
 one year ago

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Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
explain what u need to find
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \frac{dy}{dx}=2y+6\] 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.
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
yes, got that part!
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \int\limits \frac{dy}{2y+6}=\int\limits dx\]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.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
go ahead ur doing great @zepdrix
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
all I need is the steps, i understand most of it.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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,\[\large \frac{1}{2}\ln2y+6=x+c\]
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
I already got that far. Now to put in the form y=9e^(2x2)3.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Multiply both sides by 2,\[\large \ln2y+6=2x+c\] 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`.\[\huge e^{\ln2y+6}=e^{2x+c}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
The exponential and the logarithm are `inverse` operations of one another, so on the left they will simply "undo" one another.\[\huge 2y+6=e^{2x+c}\]
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
oh, i get it!
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Understand how to solve for C from here? :)
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
i forgot that c was a constant.
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
and could be treated like that. now it's easy i think.
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
but what about the 9?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Are you not coming up with a 9? :o
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
y=e^(2x+c)/23
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
There's what I've got so far.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Ok cool. You need to solve for C at some point :) That's where your 9 will come from.
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
Don't fully comprehend, could you please help a little more.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\huge (\color{royalblue}{1},\color{orangered}{6}) \qquad \rightarrow \qquad \color{orangered}{y}=Ce^{2\color{royalblue}{x}}3\]Understand how to plug these values in? :)
 one year ago

rosho Group TitleBest ResponseYou've already chosen the best response.0
I finally got it!!! Thanks a lot!
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
try solving mine plz
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
The silly bread problem? c:
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ya
 one year ago
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