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

##### 1 Answer

Dec 21, 2017

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

#### Explanation:

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#