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Solve the initial value problem. xy' = y 2 + y , y(4) = 2

Solve the initial value problem.
xy' = y2 + y , y(4) = 2

Grade:12th pass

1 Answers

Arun
25763 Points
3 years ago
Dear Bharat
 
dy/dx = y² +y /x
 
dy/y2 +y = dx/x
 
dy/y(y+1) = dx/x
 
(1/y - 1/y+1) dy = dx/x
 
On integrating
 
ln y - ln (y+1) = ln x + ln c
 
ln [y/y+1] = ln cx
 
Now
 
y /(y+1) = cx
 
Now
 
Put x = 4, y = 2
 
2/3 = 4c
 
c = 1/6
 
Hence
 
y/(y+1) = x/6

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