What is a general solution to the differential equation #y'=y/(x^2-1)#?

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Answer 1 · 2016-12-13T16:03:09

The answer is #y=Ksqrt((x-1)/(x+1))#

Rewrite the equation as

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

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

#x^2-1=(x-1)(x+1)#

#1/(x^2-1)=A/(x-1)+B/(x+1)#

#=(A(x+1)+B(x-1))/((x-1)x+1())#

#:. 1=A(x+1)+B(x-1)#

Let #x=1#, #=>#, #1=2A#, #=>#, #A=1/2#

Let #x=-1#, #=>#, #1=-2B#, #=>#, #B=-1/2#

So,

#1/(x^2-1)=(1/2)/(x-1)-(1/2)/(x+1)#

#:.dy/y=dx/(x^2-1)=(dx/2)/(x-1)-(dx/2)/(x+1)#

#intdy/y=int(dx/2)/(x-1)-int(dx/2)/(x+1)#

#lny=ln(x-1)/2-ln(x+1)/2+C#

#lny=1/2(ln(x-1)-ln(x+1))+C#

#lny=1/2ln((x-1)/(x+1))+C#

#y=Ksqrt((x-1)/(x+1))#