Completely stuck on what I think is implicit differentiation?

[Jasonsdonervan]

Hi guys! Thanks for your help, I'm completely stuck on this question.

Thanks, Jason

 

[tkhunny]

Not sure what you could be missing.

It appears to be a separable differential equation. Separate it and apply some antiderivatives.

See where it leads.

 

[dunkelheit]

It is a differential equation, since by hypothesis [MATH]y>0[/MATH] you have that [MATH]y(x)=0[/MATH] is not a solution; so you can divide both sides by [MATH]xy(x+2)[/MATH] (since by hypothesis [MATH]x>0[/MATH] so [MATH]x(x+2) \ne 0[/MATH]) and integrate both sides in the interval [MATH][2,x][/MATH]

[MATH]\int_2^x \frac{y'(s)}{y(s)} \text{d}s=\int_2^x \frac{1}{s(s+2)} \text{d}s[/MATH]​

Can you go on from here?

 

[Jasonsdonervan]

Thank you for your replies! I continued it as a differential equation and got this? Does this seem to answer the question?

 

[Jasonsdonervan]

It is a differential equation, since by hypothesis [MATH]y>0[/MATH] you have that [MATH]y(x)=0[/MATH] is not a solution; so you can divide both sides by [MATH]xy(x+2)[/MATH] (since by hypothesis [MATH]x>0[/MATH] so [MATH]x(x+2) \ne 0[/MATH]) and integrate both sides in the interval [MATH][2,x][/MATH]

[MATH]\int_2^x \frac{y'(s)}{y(s)} \text{d}s=\int_2^x \frac{1}{s(s+2)} \text{d}s[/MATH]​

Can you go on from here?

Thank you for your reply! I continued it as a differential equation and got this? Is this the correct answer?

 

[dunkelheit]

You're welcome! The integral of [MATH]\frac{1}{x(x+2)}[/MATH] is wrong, since [MATH]1[/MATH] is not the derivative of [MATH]x(x+2)[/MATH] it is wrong that [MATH]\int \frac{1}{x(x+2)} \text{d}x=\ln [(x(x+2)]+c[/MATH].

 

[Jasonsdonervan]

You're welcome! The integral of [MATH]\frac{1}{x(x+2)}[/MATH] is wrong, since [MATH]1[/MATH] is not the derivative of [MATH]x(x+2)[/MATH] it is wrong that [MATH]\int \frac{1}{x(x+2)} \text{d}x=\ln [(x(x+2)]+c[/MATH].
Another mistake is when you write [MATH]e^c e^{\ln [x(x+2)]}=e^c x e^{\ln (x+2)}[/MATH], be careful.

Ah no okay! So, what should the integral of 1/x(x+2) be instead? And then I think that will put me on the right track. Thanks for all your help

 

[dunkelheit]

Actually I think that you shouldn't approach differential equations if you haven't assimilated integrals very well, it has no sense; that integral is very basic and I think it is only a damage for you to tackle differential equations without being fluent with integrals. I don't want to sound rude, is just that this is the only real help I can give you; so go study integrals!

 

[Steven G]

Although arriving at ln(y) = ln[x(x+2)] + c was WRONG what you did after that was not correct either. It is very worthwhile to go over your error.

Note that if ln(A) = ln(b) then A=B.

In ln(y) = ln[x(x+2)] + c I would view the rhs as follows. ln[x(x+2)] + c = ln[x(x+2)] + ln(c) = ln[cx(x+2)]

So if ln(y) = ln[cx(x+2)] we have y = cx(x+2)

For the integral try partial fractions!

 

[HallsofIvy]

I was a little surprised that you recognize that \(\displaystyle e^{ln(y)}= y\) but did not recognize that \(\displaystyle e^{ln(x(x+2)}= x(x+2)\)!

However, as dunkelheit told you, that is wrong because you integrated incorrectly. To integrate \(\displaystyle \int \frac{dx}{x(x+2)}\) use "partial fractions". Find A and B such that \(\displaystyle \frac{1}{x(x+2)}= \frac{A}{x}+ \frac{B}{x+2}\).