Help! Differential Equations - C4 Watch

clear.twilight
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Ok, this is totally knocking my confidence!

I have done the first question of my homework and it appears to be wrong... However, our textbooks do have the wrong answer occasionally, so i'm hoping this is one of those cases.

Find the general solution for:

\frac{dy}{dx} = \frac{x + 1}{y}

So this is what I did...

y \frac{dy}{dx} = x + 1

\displaystyle\int  y \, dy = \displaystyle\int x + 1 \, dx

\frac{1}{2}y^{2} = \frac{1}{2}x^{2} + x + c

y^{2} = x^{2} + 2x + c

y = \pm \sqrt {x^{2} + 2x + c}

However the answer in the back is:

y = \pm \sqrt {(x + 1)^{2} + c}

I dont see how this could be the case... Where would the +1 come from to get that factorisation?

Any help please?
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generalebriety
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(Original post by clear.twilight)
Ok, this is totally knocking my confidence!

I have done the first question of my homework and it appears to be wrong... However, our textbooks do have the wrong answer occasionally, so i'm hoping this is one of those cases.

Find the general solution for:

\frac{dy}{dx} = \frac{x + 1}{y}

So this is what I did...

y \frac{dy}{dx} = x + 1

\displaystyle\int  y \, dy = \displaystyle\int x + 1 \, dx

\frac{1}{2}y^{2} = \frac{1}{2}x^{2} + x + c

y^{2} = x^{2} + 2x + c

y = \pm \sqrt {x^{2} + 2x + c}

However the answer in the back is:

y = \pm \sqrt {(x + 1)^{2} + c}


Any help please?
The two are equivalent. (Put "your c" = 1 + "their c".)

An arbitrary constant is just that - arbitrary. Adding 1 to it makes it no less arbitrary. They've decided, for some reason, that it looks more fun to write it like that. Actually, more likely is that they just integrated (x+1) directly to 1/2(x+1)^2 + const.
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clear.twilight
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Is there a way I could have got their answer though?

I seeeeee, so rather than splitting the x + 1 up they have just integrated it as a bracket? Which way is best to do...? Would I still get the marks in an a level exam doing it how I did?
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Glutamic Acid
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It's the same, it just differs by an arbitrary constant. If you expanding (x+1)^2 you'll get x^2+2x+1+c, so you could just say d = 1 + c. Sorry, that was an awfully convoluted explanation.
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generalebriety
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(Original post by clear.twilight)
Is there a way I could have got their answer though?
See the edit to my post.

Both are right, and there's no reason to say one is better than another...
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qgujxj39
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If you put your answer in an exam, it would get full marks.
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clear.twilight
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Ok, thanks

So basically if my answer differs from theres by a constant its still right? As long as the x's and y's are right it doesnt matter whether there is -3 + c or just +c for example?
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qgujxj39
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Yep.

Another example is if your answer includes a logarithm. If you have the general solution
y = ln(x) + C
then
y = ln(Cx) is equally valid, as you could've written the constant as ln(C) and then used the log rules to add it to ln(x).
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clear.twilight
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Oooo, ok. Thanks
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generalebriety
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While we're doing this, I might as well add a small but important point (see it as a piece of pedantry in this example, but it can be important in others):

(Original post by clear.twilight)
\frac{1}{2}y^{2} = \frac{1}{2}x^{2} + x + c

y^{2} = x^{2} + 2x + c
This step is, of course, completely wrong. You're basically saying that c = 2c. It's generally 'good style' to rename the constant if you're going to do this. If you're gonna be making lots of little changes to it and you don't want to keep changing the name of the constant, you can just write "+ const." or "+ constant", but actually giving it a letter c implies that when you multiply the whole equation by 2, that constant will change to 2c.

This is obviously important in places where you might manipulate the second expression to work out c and then try and plug it back into the first expression for some reason. Not an issue here, but something you need to watch out for.
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clear.twilight
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Oooo, ok. Thanks for that
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