Solve differential equation y'(t)= -3y+9, y(0)=4

Chardonnay Felix 2021-02-06 Answered
Solve differential equationy(t)=3y+9, y(0)=4
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Brittany Patton
Answered 2021-02-07 Author has 100 answers

dydt=3y+9
dydt+3y=9
I.F.=e3dt=e3t
Dt(e3ty)=9e3t
e3ty=9e3tdt
e3ty=27e3t+C
y(t)=27e3ty
now as y(0)=4
y(0)=27+Ce(0)
C= 4-27
C= -23
C= -23
y(t)=2723e3t

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Jeffrey Jordon
Answered 2021-12-25 Author has 2262 answers

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We have y = x / y, which is a first-order homogeneous differential equation.
It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that y = ± x 2 + c .
Now if we use the substitution y = u x y = u x + u ,, and rewrite the differential equation as
u x + u = 1 u
and then rearrange to
( 1 1 / u u ) d u = ( 1 x ) d x
by integrating both parts we get that
(1) 1 2 ln | u 2 1 | = ln | x | + c
For y = ± x (a special solution for c=0) u = ± 1, and by plugging ± 1 into (1) we get that
(2) ln | 0 | = ln | x | + c
What does equation (2) mean? ln | 0 | is undefined. Is this of any significance?
Edit 1:
As pointed out when rearranging from u x + u = 1 u to ( 1 1 / u u ) d u = ( 1 x ) d x, we implicitly assumed that u ± 1. Equation (1) does not hold for u = ± 1
Edit 2:
Solving equation (1) for u with u ± 1, we arrive at the same family of equations but with c 0. The fact that c can be zero comes from setting u = ± 1

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