Solve the initial value problem:

**andvaka** · June 24th 2012, 10:35 AM

(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.

**biffboy** · June 24th 2012, 11:06 AM

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.

**andvaka** · June 24th 2012, 01:51 PM

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.

**skeeter** · June 24th 2012, 02:43 PM

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 $(x+1)$

**HallsofIvy** · June 24th 2012, 06:05 PM

The typo was not by biffboy. Andvaka originally wrote "(dy/dx) = (y+1(x-1))" which is NOT "(dy/dx)= (y+1)(x-1)".

**Prove It** · June 24th 2012, 07:42 PM

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...

**andvaka** · June 24th 2012, 07:45 PM

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!

**Prove It** · June 24th 2012, 07:52 PM

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 $\displaystyle \begin{align*} \frac{dy}{dx} = y + 1(x - 1) \end{align*}$ ?

**andvaka** · June 24th 2012, 08:10 PM

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.
$\displaystyle \begin{align*} \frac{dy}{dx} = (y + 1)(x - 1) \end{align*}$

**biffboy** · June 24th 2012, 09:36 PM

Sorry. Yes it should be (x-1)

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