Question

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

Answer 1

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

Answer 2

Please, can anyone talk to me about differential equations?

Answer 3

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
\[ e^{\int P(t) dt} y = \int e^{\int P(t) dt}Q(t) dt +c\]

Answer 4

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.

Answer 5

So, I get \[1+x^2 \] as my integrating factor. My understanding is I multiply both sides by that now? So I would have \[\frac{ dy }{ dt }(1+t^2) + 2ty = 1\]

Answer 6

Im not sure I understand what happens with the left side here, I think I did something wrong.

Answer 7

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))

Answer 8

or if we pick up where you left off
\[ \frac{ dy }{ dt }(1+t^2) + 2ty = 1 \\
(1+t^2) dy + 2 t \ y\ dt = dt \]

Answer 9

so multiply both sides by dt?

Answer 10

yes, we almost always split the differentials

Answer 11

ahhhh ok, I was wondering but that makes sense, back to the basics

Answer 12

though it may not be obvious the left side is
\[ d( (1+t^2) y) \]
and you have
\[ d( (1+t^2) y)= dt \\ \int d( (1+t^2) y)= \int dt
\]

Answer 13

what happens to the 2tydt on the left side?

Answer 14

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)

Answer 15

in other words if you expand d ( (1+t^2) y) you get
(1+t^2) dy + y 2t dt

Answer 16

arg ok, I think I get that. I'll play around with them until it sinks in.

Answer 17

\[ \int d( (1+t^2) y)= \int dt \\ (1+t^2) y = t+C \\ y= \frac{t}{1+t^2}+C'
\]
where we can rename C/(1+t^2) as a new constant of integration C'