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Answer 1 · 2017-06-07T08:26:30

#x(y) = (2y^5+C)/y^2#

The equation is in the form:

#dx/dy + f(y)x = g(y)#

where:

#f(y) = 2/y#

#g(y) = 10y^2#

The integrating factor is:

#A(y) = e^(int f(y)dy) = e^(int 2/ydy) = e^(2ln abs y +C) = cy^2#

The solution of the equation is then:

#x(y) = 1/(A(y)) int A(u) g(u) du#

#x(y) = 1/(cy^2) int cu^2 10u^2dy = 10/y^2 int u^4du = 2y^3+C/y^2#

In fact:

#d/dy (2y^3+C/y^2) +2/y (2y^3+C/y^2) = 6y^2-(2C)/y^3 +4y^2+(2C)/y^3 = 10y^2#