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

This Question is Closed

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

zepdrix
 one year ago
Best 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.

zepdrix
 one year ago
Best 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.

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Multiply 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`.\[\huge e^{\ln2y+6}=e^{2x+c}\]

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

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

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

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2At 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.

zepdrix
 one year ago
Best 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? :)

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

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

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