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

1. Best_Mathematician

explain what u need to find

2. zepdrix

$\large \frac{dy}{dx}=2y+6$ 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.

3. rosho

yes, got that part!

4. zepdrix

$\large \int\limits \frac{dy}{2y+6}=\int\limits dx$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.

5. Best_Mathematician

go ahead ur doing great @zepdrix

6. rosho

all I need is the steps, i understand most of it.

7. zepdrix

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,$\large \frac{1}{2}\ln|2y+6|=x+c$

8. rosho

I already got that far. Now to put in the form y=9e^(2x-2)-3.

9. zepdrix

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.$\huge e^{\ln|2y+6|}=e^{2x+c}$

10. zepdrix

The exponential and the logarithm are inverse operations of one another, so on the left they will simply "undo" one another.$\huge 2y+6=e^{2x+c}$

11. rosho

oh, i get it!

12. zepdrix

Understand how to solve for C from here? :)

13. rosho

i forgot that c was a constant.

14. rosho

and could be treated like that. now it's easy i think.

15. rosho

16. zepdrix

Are you not coming up with a 9? :o

17. rosho

y=e^(2x+c)/2-3

18. rosho

There's what I've got so far.

19. zepdrix

Ok cool. You need to solve for C at some point :) That's where your 9 will come from.

20. rosho

21. zepdrix

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.

22. zepdrix

$\huge (\color{royalblue}{1},\color{orangered}{6}) \qquad \rightarrow \qquad \color{orangered}{y}=Ce^{2\color{royalblue}{x}}-3$Understand how to plug these values in? :)

23. rosho

I finally got it!!! Thanks a lot!

24. zepdrix

yay! \c:/

25. Best_Mathematician

try solving mine plz

26. zepdrix