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# Linear First Order Equation?

Please help, given is a linear first order differential equation but I dont know how to solve it (using integrating factor method)

(3x^(2)+y+3x^(3)y)dx + xdy = 0

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- Ian HLv 71 month ago
[3x^2 + y + 3x^3y]*dx + xdy = 0 ... div by xdx

3x + y/x + 3x^2y + dy/dx = 0

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

This DE solution requires that we know that ......

dy/dx + P(x)y = Q(x) has a solution given by

I(x)*y = ∫[I(x)Q(x)] dx

where I = e^∫P(x)dx is called the INTEGRATING FACTOR

In this example P = 3x^2 + 1/x, so ∫P(x)dx = x^3 + ln(x)

I = e^[ x^3 + ln(x)] = e ln(x)* e^x^3 = xe^x^3

∫[I(x)Q(x)] dx = ∫[-3x*xe^x^3] dx = ∫(-3x^2*e^x^3)dx

∫[I(x)Q(x)] dx = -e^x^3 + C, so now we have

xe^x^3*y = C - e^x^3

xy = Ce^-x^3 – 1 or,

y = [Ce^-x^3 – 1]/x

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