C4: confused differential equations

Hi,

So, I've come across a question which I thought I worked out correctly but my answer is way off so now I'm left confused. The question was:

$\frac{dy}{dx}=\frac{y}{(x+1)(x+2)}$ and y=2, x=1

So I put the y terms on the LHS and the left the x terms on the RHS. I then split the fraction on the RHS. This gave me the following to integrate:

$\displaystyle \int \frac{1}{y} dy = \displaystyle \int \frac{1}{x+1} - \frac{1}{x+2} dx$

This gave me the following:

ln|y| = ln|x+1| - ln|x+2| + c

c= ln|y| - ln|x+1| + ln|x+2|

c= ln|3| (when y=2, x=1, as given above).

Subbing this back into ln|y| = ln|x+1| - ln|x+2| + c gave me:

ln|y| = ln|x+1| - ln|x+2| + ln|3|

Raising everything to base e:

y=x+1 -(x+2) + 3
y=2

The correct answer is $y=\frac{3(x+1)}{x+2}$

When I took the approach of splitting the fraction, I was unsure if it was the correct thing to do, but I didn't know how else to approach the denominator.

Could anyone help?
Thanks
Marc

Hi,

So, I've come across a question which I thought I worked out correctly but my answer is way off so now I'm left confused. The question was:

$\frac{dy}{dx}=\frac{y}{(x+1)(x+2)}$ and y=2, x=1

So I put the y terms on the LHS and the left the x terms on the RHS. I then split the fraction on the RHS. This gave me the following to integrate:

$\displaystyle \int \frac{1}{y} dy = \displaystyle \int \frac{1}{x+1} - \frac{1}{x+2} dx$

This gave me the following:

ln|y| = ln|x+1| - ln|x+2| + c

c= ln|y| - ln|x+1| + ln|x+2|

c= ln|3| (when y=2, x=1, as given above).

Subbing this back into ln|y| = ln|x+1| - ln|x+2| + c gave me:

ln|y| = ln|x+1| - ln|x+2| + ln|3|

Raising everything to base e:

y=x+1 -(x+2) + 3
y=2

The correct answer is $y=\frac{3(x+1)}{x+2}$

When I took the approach of splitting the fraction, I was unsure if it was the correct thing to do, but I didn't know how else to approach the denominator.

Could anyone help?
Thanks
Marc

Hi,

So, I've come across a question which I thought I worked out correctly but my answer is way off so now I'm left confused. The question was:

$\frac{dy}{dx}=\frac{y}{(x+1)(x+2)}$ and y=2, x=1

So I put the y terms on the LHS and the left the x terms on the RHS. I then split the fraction on the RHS. This gave me the following to integrate:

$\displaystyle \int \frac{1}{y} dy = \displaystyle \int \frac{1}{x+1} - \frac{1}{x+2} dx$

This gave me the following:

ln|y| = ln|x+1| - ln|x+2| + c

c= ln|y| - ln|x+1| + ln|x+2|

c= ln|3| (when y=2, x=1, as given above).

Subbing this back into ln|y| = ln|x+1| - ln|x+2| + c gave me:

ln|y| = ln|x+1| - ln|x+2| + ln|3|

Raising everything to base e:

y=x+1 -(x+2) + 3
y=2

The correct answer is $y=\frac{3(x+1)}{x+2}$

When I took the approach of splitting the fraction, I was unsure if it was the correct thing to do, but I didn't know how else to approach the denominator.

Could anyone help?
Thanks
Marc

When you raise the lns, the whole of the RHS is raised to a single e, not each separate component.

Hi,

So, I've come across a question which I thought I worked out correctly but my answer is way off so now I'm left confused. The question was:

$\frac{dy}{dx}=\frac{y}{(x+1)(x+2)}$ and y=2, x=1

So I put the y terms on the LHS and the left the x terms on the RHS. I then split the fraction on the RHS. This gave me the following to integrate:

$\displaystyle \int \frac{1}{y} dy = \displaystyle \int \frac{1}{x+1} - \frac{1}{x+2} dx$

This gave me the following:

ln|y| = ln|x+1| - ln|x+2| + c

c= ln|y| - ln|x+1| + ln|x+2|

c= ln|3| (when y=2, x=1, as given above).

Subbing this back into ln|y| = ln|x+1| - ln|x+2| + c gave me:

ln|y| = ln|x+1| - ln|x+2| + ln|3|

When you exponentiate this, you do not get what is written below in blue. You would get:

$e^{ln|x-1|-ln|x+2|+ln|3|}=e^{ln|\frac{3(x-1)}{x+2}|}[br]$

Raising everything to base e:

y=x+1 -(x+2) + 3
y=2

The correct answer is $y=\frac{3(x+1)}{x+2}$

When I took the approach of splitting the fraction, I was unsure if it was the correct thing to do, but I didn't know how else to approach the denominator.

Could anyone help?
Thanks
Marc

See above.

Thanks! But after you've exponentiated it, where does the ln|x-1| come from, and where has the ln|y| gone?

Typo, I mean ln|x+1|

The ln|y| is now y. I haven't written it down though, I was explaining why your RHS is incorrect.

Do you understand now?

Yes, I do, perfectly

Thanks you for your time and help.

I used the log laws and solved it. I know the whole thing gets raised to the base e, but how would that look as opposed to what I did, because that is what I tried to do.