# Solve the initial value problem:

• Jun 24th 2012, 10:35 AM
andvaka
Solve the initial value problem:
(dy/dx) = (y+1(x-1)) ; y(1)=3 variable separable

We've only recently began the unit on Integration applications and Differentials, so I have absolutely no idea how to approach this. If any of you can show me how to solve this, it would be most appreciated.
• Jun 24th 2012, 11:06 AM
biffboy
Re: Solve the initial value problem:
Quote:

Originally Posted by andvaka
(dy/dx) = (y+1(x-1)) ; y(1)=3 variable separable

We've only recently began the unit on Integration applications and Differentials, so I have absolutely no idea how to approach this. If any of you can show me how to solve this, it would be most appreciated.

Separating the variables we get dy/(y+1)=(x+1)dx Integrate both sides log (y+1)= x^2/2+x+C
When x=1 y=3 so log4=(1/2)+1+C So C=log4-3/2 etc.
• Jun 24th 2012, 01:51 PM
andvaka
Re: Solve the initial value problem:
Quote:

Originally Posted by biffboy
Separating the variables we get dy/(y+1)=(x+1)dx Integrate both sides log (y+1)= x^2/2+x+C
When x=1 y=3 so log4=(1/2)+1+C So C=log4-3/2 etc.

Shouldn't the (x+1)dx in "dy/(y+1)=(x+1)dx" be (x-1)dx . or is there a reason for the sign change?
Thanks.
• Jun 24th 2012, 02:43 PM
skeeter
Re: Solve the initial value problem:
Quote:

Originally Posted by andvaka
(dy/dx) = (y+1(x-1)) ; y(1)=3 variable separable

We've only recently began the unit on Integration applications and Differentials, so I have absolutely no idea how to approach this. If any of you can show me how to solve this, it would be most appreciated.

if the DE is

$\frac{dy}{dx} = (y+1)(x-1)$

then

$\int \frac{dy}{y+1} = \int (x-1) \, dx$

... most probably a typo by biff to put $(x+1)$
• Jun 24th 2012, 06:05 PM
HallsofIvy
Re: Solve the initial value problem:
The typo was not by biffboy. Andvaka originally wrote "(dy/dx) = (y+1(x-1))" which is NOT "(dy/dx)= (y+1)(x-1)".
• Jun 24th 2012, 07:42 PM
Prove It
Re: Solve the initial value problem:
If the original function actually is as written, then it is \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}, which would need to be solved using an Integrating Factor...
• Jun 24th 2012, 07:45 PM
andvaka
Re: Solve the initial value problem:
Quote:

Originally Posted by Prove It
If the original function actually is as written, then it is \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}, which would need to be solved using an Integrating Factor...

Would you mind elaborating?
Thanks!
• Jun 24th 2012, 07:52 PM
Prove It
Re: Solve the initial value problem:
Quote:

Originally Posted by andvaka
Would you mind elaborating?
Thanks!

I would, considering I have no idea if this is the actual problem. As the other posters have stated, it appears that you had tried to write \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} with a typo, thereby making an entirely different problem. The reason it looks wrong is because you would not usually write \displaystyle \begin{align*} 1(x - 1) \end{align*}.

Which is the correct problem? \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} or \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}?
• Jun 24th 2012, 08:10 PM
andvaka
Re: Solve the initial value problem:
Quote:

Originally Posted by Prove It
I would, considering I have no idea if this is the actual problem. As the other posters have stated, it appears that you had tried to write \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} with a typo, thereby making an entirely different problem. The reason it looks wrong is because you would not usually write \displaystyle \begin{align*} 1(x - 1) \end{align*}.

Which is the correct problem? \displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*} or \displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}?

Ah. It was a typo. The first is correct.
\displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*}
• Jun 24th 2012, 09:36 PM
biffboy
Re: Solve the initial value problem:
Sorry. Yes it should be (x-1)