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Two different methods give two different answer

I was solving a differential equation and got to the step where I had to integrate. I did it two different methods to get two different answers. I was wondering if anyone knows why one method must not work.

Method 1

∫14ydy \int \frac{1}{4y} dy

let u=4y u = 4y

therefore:

dudy=4 \frac{du}{dy} = 4

dydu=14 \frac{dy}{du} = \frac{1}{4}

dy=14du dy = \frac{1}{4} du

so:

∫1u14du=14∫1udu \int \frac{1}{u} \frac{1}{4} du = \frac{1}{4} \int \frac{1}{u} du

14ln⁡u=14ln⁡(4y) \frac{1}{4} \ln u = \frac{1} {4} \ln (4y)

Method 2

∫14ydy \int \frac{1}{4y} dy

14∫1y \frac{1}{4} \int \frac{1}{y}

14ln⁡y \frac{1}{4} \ln y

As you see both gives different answer but I'm sure both methods are correct.

Thanks,
Kieron
You missed off the '+c'. Your first answer is equivalent to 1/4*ln(4) + 1/4*ln(y).
Reply 2
Avatar for nuodai
nuodai
17
The key is the "+C", which you forgot to add.

Notice that if g(x)=f(x)+Cg(x) = f(x) + C then dgdx=dfdx\dfrac{dg}{dx} = \dfrac{df}{dx}, so when you integrate two things then you can end up with two different answers, but they will always only differ by a constant.

Here, ln⁡4y=ln⁡4+lny\ln 4y = \ln 4 + ln y, so you get 14ln⁡4+14ln⁡y\dfrac{1}{4} \ln 4 + \dfrac{1}{4} \ln y. The 14ln⁡4\dfrac{1}{4}\ln 4 is a constant, and so (if you add your constants of integration) the two answers are actually the same -- this 14ln⁡4\dfrac{1}{4}\ln 4 can then be absorbed into the constant of integration in your first method to give 14ln⁡y+C\dfrac{1}{4}\ln y + C.
Reply 3
Avatar for Hopple
Hopple
18
You've forgotten your constant of integration.
Original post by nuodai
The key is the "+C", which you forgot to add.

Notice that if g(x)=f(x)+Cg(x) = f(x) + C then dgdx=dfdx\dfrac{dg}{dx} = \dfrac{df}{dx}, so when you integrate two things then you can end up with two different answers, but they will always only differ by a constant.

Here, ln⁡4y=ln⁡4+lny\ln 4y = \ln 4 + ln y, so you get 14ln⁡4+14ln⁡y\dfrac{1}{4} \ln 4 + \dfrac{1}{4} \ln y. The 14ln⁡4\dfrac{1}{4}\ln 4 is a constant, and so (if you add your constants of integration) the two answers are actually the same -- this 14ln⁡4\dfrac{1}{4}\ln 4 can then be absorbed into the constant of integration in your first method to give 14ln⁡y+C\dfrac{1}{4}\ln y + C.


I can't believe something so simple drove me crazy. Most of the times the constant doesn't change the answer so I tend to forget to put it.

Thanks :biggrin:
Reply 5
Avatar for Jodin
Jodin
2
Original post by Darkening Light
I can't believe something so simple drove me crazy. Most of the times the constant doesn't change the answer so I tend to forget to put it.

Thanks :biggrin:


Indeed! In regards to forgetting about the constant, try integrating 1/x dx by parts some time, with u = 1/x, v' = 1.
Reply 6
Avatar for pi+e=pie
pi+e=pie
the +c my friend!
i forgot it when integrating arcsin and i got two different answers by two methods aswell.
not to worry, just an easy mistake

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