### Author Topic: TUT0702 quiz and solutions  (Read 613 times)

#### Zhangxinbei

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« on: September 27, 2019, 04:39:52 PM »
Find the initial value problem.  Y' = 2x/1+y^2, y(2)=0.
Using separable:
dy/dx = (2x)/(1+y^2)
integral on both sides:
∫ 1+y^2 dy=∫ 2x dx
y+(y^3)/3 = x^2 + c
as y(2) = 0, plug into the function:
0 = 4+c
c = -4
Therefore,  y+(y^3)/3 = x^2 -4

#### Yan

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« Reply #1 on: September 30, 2019, 09:48:59 PM »
Hi, I find your answer to that question is very helpful, and I have annother similar question that you can try to answer.

Find the general solution of the given differential equation, and use it to determine how solutions as t approaches infinity.

ty' + 2y = sin(t), t>0.

#### Zhangxinbei

• Jr. Member
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• Karma: 0 ##### Re: TUT0702 quiz and solutions
« Reply #2 on: October 04, 2019, 03:20:52 PM »
Sure!
y'+ 2/t y = sint/t
y' + p(t)y = g(t)
p(t) = 2/t.  G(t) = sint/t
u = e^∫2/t dt = t^2
multiply t^2 to both sides:
t^2y' + 2ty = t sint
(t^2y)' = t sint
t^2y = ∫t sint dt
y = (-t cost + pint +c)/t^2

SO, given t>0, y->o, as t -> infinity