Asked by Taldil
Please help me solve this differential equation: dy/dx-(1+1/x)y=y^2
Answers
Answered by
Steve
This is just a special case of the Bernoulli Equation
y' + p(x) y = q(x) y^n
where q(x) = 1
Make the substitution v = y^(1-n)
This will give you
-v' + (1+1/x)v = 1
Now you can multiply by an integrating factor
e^[∫(1+1/x) dx] = xe^x to get
xe^x v = -e^x (x-1)
and you're almost done
y' + p(x) y = q(x) y^n
where q(x) = 1
Make the substitution v = y^(1-n)
This will give you
-v' + (1+1/x)v = 1
Now you can multiply by an integrating factor
e^[∫(1+1/x) dx] = xe^x to get
xe^x v = -e^x (x-1)
and you're almost done
Answered by
Taldil
Hi..
I'm not sure how to get the final answer
I'm not sure how to get the final answer
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