Find the general solution of the given differential equation

[shamieh]

Find the general solution of the given differential equation..

$$(1+t^2)y' + 4ty = (1+t^2)^{-2}$$

I'm kind of confused here on what to do...

Do I want to do something like $$e ^{\int4t} dt$$ and then multiply that through on both sides or do I need to do something different here..I'm not really sure how to approach this one so that's why I'm kind of beating around the bush if you will.

 

[MarkFL]

I would first divide through by $$1+t^2$$ to write the ODE in standard linear form, and observe that in doing so no trivial solutions are being lost:

$$\d{y}{t}+\frac{4t}{t^2+1}y=\left(t^2+1\right)^{-3}$$

Next, compute your integrating factor:

$$\mu(t)=\exp\left(\int\frac{4t}{t^2+1}\,dt\right)$$

What do you get for this factor?

 

[shamieh]

$$mew = 2\ln(t^2 + 1)$$ correct?

 

[MarkFL]

$$mew = 2ln(t^2 + 1)$$ correct?

Not quite...and use the code \mu for the Greek letter $\mu$...you should get:

$$\mu(t)=\exp\left(2\ln\left(t^2+1\right)\right)=\exp\left(\ln\left(\left(t^2+1\right)^2\right)\right)=\left(t^2+1\right)^2$$

Now, applying this factor, what is your ODE? If we computed the factor correctly, the left side of the ODE will be the differentiation of a product...can you make this transformation?

 

[shamieh]

so the left side will be $$(t^2+1)^2y $$ correct?

 

[MarkFL]

so the left side will be $$(t^2+1)^2y $$ correct?

No, the left side will be:

$$\left(t^2+1\right)^2\d{y}{t}+4t\left(t^2+1\right)y=\frac{d}{dt}\left(\left(t^2+1\right)^2y\right)$$

Now your ODE is:

$$\frac{d}{dt}\left(\left(t^2+1\right)^2y\right)=\frac{1}{t^2+1}$$

So just integrate both sides with respect to $t$...what do you get?

 

[shamieh]

$$y = \frac{\arctan(t) + c}{(1+t^2)^2}$$

 

[MarkFL]

$$y = \frac{arctan(t) + c}{(1+t^2)^2}$$

Correct! (Yes)

Another $\LaTeX$ tip:

Precede functions, like logarithms, trig functions and their inverses, and hyperbolic functions, etc with a backslash. This will prevent them from being italicized and emphasize that they are functions rather than a string of variables. :D

 

[shamieh]

Thanks! Will do. Haven't been on the forum in some time and have forgotten LaTeX. I just started Calculus IV today and it's pretty interesting. I'll be on here for Discrete Mathematics help as well. Site looks awesome since I was last here btw.

 

[MarkFL]

Thanks! Will do. Haven't been on the forum in some time and have forgotten LaTeX. I just started Calculus IV today and it's pretty interesting. I'll be on here for Discrete Mathematics help as well. Site looks awesome since I was last here btw.

Yes, we've been pretty busy making changes...server upgrade, PHP upgrade, vBulletin software upgrades, thanks system upgrade, and a bunch of styling changes...I am happy all of our work is noticeable. :D