Asked by Chintu
Initial Value Problem
dy/dx = y^2 -5y+6
y(0)=1
dy/dx = y^2 -5y+6
y(0)=1
Answers
Answered by
Steve
y' = y^2-5y+6
dy/((y-3)(y-2)) = dx
dy (1/(y-3) - 1/(y-2)) = dx
ln[(y-3)/y-2)] = x + c
(y-3)/(y-2) = c e^x
y(0) = 1, so
(1-3)/(1-2) = c
c = 2
(y-3)/(y-2) = 2e^x
Now you can rearrange things and find y as a function of x:
y = (3-4e^x)/(1-2e^x)
dy/((y-3)(y-2)) = dx
dy (1/(y-3) - 1/(y-2)) = dx
ln[(y-3)/y-2)] = x + c
(y-3)/(y-2) = c e^x
y(0) = 1, so
(1-3)/(1-2) = c
c = 2
(y-3)/(y-2) = 2e^x
Now you can rearrange things and find y as a function of x:
y = (3-4e^x)/(1-2e^x)
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