Solve the Differential Equation # dy/dx = 2xy - x #?

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Answer 1 · 2017-12-21T12:44:51

# y = Ae^(x^2) + 1/2 #

We have:

# dy/dx = 2xy - x #

We can factorize the RHS to get

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

Which is separable, so we can "separate the variables" to get:

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

So we integrate

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

And we can rearrange:

# ln(2y-1) = x^2 + 2C #
# :. 2y-1 = e^(x^2 + 2C) #
# :. 2y-1 = e^(x^2)e^(2C) #
# :. y = Ae^(x^2) + 1/2#