anonymous
  • anonymous
Missing something simple! Check my work so far please? Diff EQs. The problem is \[\frac{ dy }{ dt } + \frac{ 2t }{ 1+t^2 }y = \frac{ 1 }{ 1+t^2 }\] So far, I move all terms to the left so: \[\frac{ dy }{ dt } + \frac{ 2t }{ 1+t^2 }y - \frac{ 1 }{ 1+t^2 } =0\] Next I divided both sides by y, giving : \[\frac{ \frac{ dy }{ dt } }{ y } + \frac{ 2t }{ 1+t^2 } - \frac{ 1 }{ y+yt^2 } = 0\] and now I am stuck
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
I have the identity that \[\frac{ \frac{ dy }{ dt } }{ y } = \frac{ d }{ dt }\ln \left| y(t) \right|\] but I feel like I am missing a step before exponentiating each side
anonymous
  • anonymous
Please, can anyone talk to me about differential equations?
phi
  • phi
Do you know about Integrating Factors? a linear differential equation of the form \[ \frac{dy}{dt} +P(t) y = Q(t) \] has an integrating factor \[ e^{\int P(t) dt} \] the solution will be \[ e^{\int P(t) dt} y = \int e^{\int P(t) dt}Q(t) dt +c\]

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anonymous
  • anonymous
i just watched a video on them and had my mind blown. Thanks so much for the response, glad to know Im finally on the right track.
anonymous
  • anonymous
So, I get \[1+x^2 \] as my integrating factor. My understanding is I multiply both sides by that now? So I would have \[\frac{ dy }{ dt }(1+t^2) + 2ty = 1\]
anonymous
  • anonymous
Im not sure I understand what happens with the left side here, I think I did something wrong.
phi
  • phi
I would start out as \[ \frac{ dy }{ dt } + \frac{ 2t }{ 1+t^2 }y = \frac{ 1 }{ 1+t^2 } \\ dy +\frac{ 2t }{ 1+t^2 }y \ dt= \frac{ 1 }{ 1+t^2 } \ dt \] now multiply by the integrating factor \[ (1+t^2) dy +(1+t^2)\frac{ 2t }{ 1+t^2 }y \ dt= (1+t^2)\frac{ 1 }{ 1+t^2 } \ dt \\ (1+t^2) dy + 2t\ y\ dt = dt \] now integrate both sides. the left side is d (f(t) y) = f(t) dy + y d(f(t))
phi
  • phi
or if we pick up where you left off \[ \frac{ dy }{ dt }(1+t^2) + 2ty = 1 \\ (1+t^2) dy + 2 t \ y\ dt = dt \]
anonymous
  • anonymous
so multiply both sides by dt?
phi
  • phi
yes, we almost always split the differentials
anonymous
  • anonymous
ahhhh ok, I was wondering but that makes sense, back to the basics
phi
  • phi
though it may not be obvious the left side is \[ d( (1+t^2) y) \] and you have \[ d( (1+t^2) y)= dt \\ \int d( (1+t^2) y)= \int dt \]
anonymous
  • anonymous
what happens to the 2tydt on the left side?
phi
  • phi
the product rule (take the derivative of the product of two variables) is d (u v) = u dv + v du in this problem we go in the other direction u dv + v du = d (u v)
phi
  • phi
in other words if you expand d ( (1+t^2) y) you get (1+t^2) dy + y 2t dt
anonymous
  • anonymous
arg ok, I think I get that. I'll play around with them until it sinks in.
phi
  • phi
\[ \int d( (1+t^2) y)= \int dt \\ (1+t^2) y = t+C \\ y= \frac{t}{1+t^2}+C' \] where we can rename C/(1+t^2) as a new constant of integration C'

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