Question #866a2

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Answer 1 · 2017-03-20T19:12:27

# y = 3 - 4/x^2#

This is separable:

#1/(3 - y )dy/dx = 2/x#

Integrate wrt x:

#int 1/(3 - y )dy/dx \ dx =int 2/x \ dx#

By chain rule:

#int 1/(3 - y ) \ dy =int 2/x \ dx#

# - ln (3 - y ) = 2 ln x + C#

Apply the #(2,2)# IV:

# - ln (1 ) = 2 ln 2 + C#

# implies C = - 2 ln 2 = ln (1/4)#

So
# - ln (3 - y ) = ln x^2 + ln(1/ 4) #

# ln (1/(3 - y )) = ln (x^2/4)#

# 3 - y = 4/x^2#

# y = 3 - 4/x^2#