MAT244--2019F > Quiz-1

TUT0702 quiz and solutions

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**Zhangxinbei**:

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**:

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**:

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

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