Resolve the differential equation : xy' + 4 = y^2 with the condition : y(1)=1
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Resolve the differential equation : xy' + 4 = y^2 with the condition : y(1)=1

[From: ] [author: ] [Date: 11-06-21] [Hit: ]
......

Sub in the point (1, 1):

(1/4) ln|1 - 2| - (1/4) ln|1 + 2| = ln|1| + C
(1/4) ln(1) - (1/4) ln(3) = ln(1) + C
(-1/4)ln(3) = C
ln( 3^(-1/4) ) = C

(1/4) ln|y - 2| - (1/4) ln|y + 2| = ln|x| + ln( 3^(-1/4) )
(1/4) [ln|y - 2| - ln|y + 2|] = ln|3^(-1/4) x|
(1/4) [ln|(y - 2) / (y + 2)|] = ln|3^(-1/4) x|
ln|{ (y - 2) / (y + 2) }^(1/4)| = ln|3^(-1/4) x|
[ | (y - 2) / (y + 2) | ]^(1/4) = | 3^(-1/4) x |
| (y - 2) / (y + 2) | = (1/3)x^4

Now there are a couple ways to isolate y because of the absolute value brackets, either:

y = (6 - 2x^4) / (x^4 + 3)
y = (6 + 2x^4) / (x^4 - 3)

But we need (1, 1) to be a point, which means that we must have the first one:

y = (6 - 2x^4) / (x^4 + 3)

Done!

-
y²dx=xdy+4dx
(y²-4)dx=xdy
dx/x=dy/(y²-4)
ln(x)=asin(y/2)+c
c=-asin(1/2)
=-0.524 rad
=-30°
we can express the equation as
x=e^[asin(y/2)-30°] or

y=sin[ln(x)+30°]

-
simplified:
2(x-1)/(1-x)
-2(1-x)/(1-x)
y = -2
12
keywords: condition,with,differential,039,Resolve,xy,equation,the,Resolve the differential equation : xy' + 4 = y^2 with the condition : y(1)=1
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