# Explanation of Solution Help

• Jan 28th 2014, 03:41 PM
petenice
Explanation of Solution Help
My online class provided this solution but i have questions about some of the steps. My online professor has been no help. Any input is appreciated.

Question Solve the initial value problem. $\frac{dy}{dx}=\frac{y-1}{x+3}$

Step 1 Separate and integrate $\int\frac{dy}{y-1}=\int\frac{1}{x+3}dx$

Step 2 Integration result $ln(y-1)=ln(x+3)+lnC$ So where did the $lnC$ come from. I know where the $C$ came from but why are they multiplying it by $ln$

Step 3 Raise all to the e $e^{ln(y-1)}=e^{ln(x+3)}+e^{lnC}$

Step 4 Result is $(y-1)=C(x+3)$ Why are they multiplying the right side by $C$ shouldnt it be like this $(y-1)=(x+3)+C$

Step 5 $(y-1)=-C(x+3)$ why it the right side now negative?

Step 6 $y=1-C(x+3)$ Solution given to us by the professor.

I cant figure out how they arrived to some of those steps maybe someone here can. Thanks.
• Jan 28th 2014, 04:22 PM
romsek
Re: Explanation of Solution Help
Quote:

Originally Posted by petenice
My online class provided this solution but i have questions about some of the steps. My online professor has been no help. Any input is appreciated.

Question Solve the initial value problem. $\frac{dy}{dx}=\frac{y-1}{x+3}$

Step 1 Separate and integrate $\int\frac{dy}{y-1}=\int\frac{1}{x+3}dx$

Step 2 Integration result $ln(y-1)=ln(x+3)+lnC$ So where did the $lnC$ come from. I know where the $C$ came from but why are they multiplying it by $ln$

Step 3 Raise all to the e $e^{ln(y-1)}=e^{ln(x+3)}+e^{lnC}$

Step 4 Result is $(y-1)=C(x+3)$ Why are they multiplying the right side by $C$ shouldnt it be like this $(y-1)=(x+3)+C$

Step 5 $(y-1)=-C(x+3)$ why it the right side now negative?

Step 6 $y=1-C(x+3)$ Solution given to us by the professor.

I cant figure out how they arrived to some of those steps maybe someone here can. Thanks.

this is a mess. start from step 2 with a small change

$\ln(y-1)=\ln(x+3)+C_1 where C_1 \in \mathbb{R}$

now take the exponential on both sides

$y-1=e^{ln(x+3)+C_1}=e^{C_1}(x+3)$

$y=1+e^{C_1}(x+3)$

now if you like you can choose $C=\ln(\left|C_1\right|)$ and the resulting equation is

$y=1+C(x+3)$

note that C can be any number so it doesn't matter whether you use plus or minus in front of it so

$y=1-C(x+3)$ is equally valid.
• Jan 28th 2014, 10:39 PM
Prove It
Re: Explanation of Solution Help
Quote:

Originally Posted by petenice
Step 2 Integration result $ln(y-1)=ln(x+3)+lnC$ So where did the $lnC$ come from. I know where the $C$ came from but why are they multiplying it by $ln$

Hold on a minute. The integration result is NOT \displaystyle \begin{align*} \ln{ \left( y - 1 \right) } = \ln{ \left( x + 3 \right) } + \ln{ \left( C \right) } \end{align*}, it's \displaystyle \begin{align*} \ln{ \left| y - 1 \right| } = \ln{ \left| x + 3 \right| } + \ln{ \left( C \right) } \end{align*}.

Also, we are NOT "multiplying by ln". Do you understand what "ln" means? It is taking the natural (base e) logarithm! Now the reason they are writing it as \displaystyle \begin{align*} \ln{ \left( C \right) } \end{align*} is simply to aid in calculations, and noting that since the integration constant is ARBITRARY, and \displaystyle \begin{align*} \ln{ \left( C \right) } \end{align*} is still a constant, it is just as correct to write a logarithm constant.

Quote:

Step 3 Raise all to the e $e^{ln(y-1)}=e^{ln(x+3)}+e^{lnC}$

Step 4 Result is $(y-1)=C(x+3)$ Why are they multiplying the right side by $C$ shouldnt it be like this $(y-1)=(x+3)+C$
NO! The rules of exponentials do not work like that! What should be happening is to collect the logarithms onto one side and to simplify using logarithm laws first BEFORE exponentiating!

\displaystyle \begin{align*} \ln{ \left| y - 1 \right| } &= \ln{ \left| x + 3 \right| } + \ln{ \left( C \right) } \\ \ln{ \left| y - 1 \right| } - \ln{ \left| x + 3 \right| } &= \ln{ \left( C \right) } \\ \ln{ \left| \frac{y - 1}{x + 3} \right| } &= \ln{ \left( C \right) } \end{align*}

Now exponentiate.

The solutions given by your professor are VERY sloppy. Feel free to tell him/her that. In my opinion, you need to get the hang of trying a few steps out yourself without relying too much on following step-by-step solutions given by others. You know when a step is valid or not valid, and there's no reason to believe that when you have done all valid steps, that your answer, while it may look different to your professor's or others', is not correct.
• Jan 28th 2014, 11:16 PM
LimpSpider
Re: Explanation of Solution Help
Agree with the sloppy Professor statement.
• Feb 3rd 2014, 09:30 PM
petenice
Re: Explanation of Solution Help
Thanks guys. Prove It, i appreciate the breakdown. I just wanted to make sure I wasn't missing something. So far this professor has been less than helpful, but i appreciate everyones help. :)
• Feb 4th 2014, 07:45 AM
HallsofIvy
Re: Explanation of Solution Help
Quote:

Originally Posted by LimpSpider
Agree with the sloppy Professor statement.

But we don't know how much of that was the Professor and how much was petenice's memory of it.