What is the general solution of the differential equation dy/dx- 2xy = x ?

1 Answer
Sep 19, 2017

y = 3/2e^(x^2 - 1) - 1/2

Explanation:

We have:

dy/dx- 2xy = x

Which we can write as:

dy/dx = 2xy+ x
:. dy/dx = (2y+ 1)x
:. 1/(2y+ 1) dy/dx = x

Which is a first order separable differential equation, so we can "separate the variables" to get:

int \ 1/(2y+ 1) \ dy = int \ x \ dx

Integrating we get, the General Solution:

1/2ln|2y+1| = 1/2x^2 + C

Applying the initial condition y(1)=1 we find:

1/2ln3 = 1/2 + C => C = 1/2ln3 - 1/2

So we can write an implicit particular solution as:

1/2ln|2y+1| = 1/2x^2 + 1/2ln3 - 1/2

We typically require an explicit solution, so we can rearrange as follows:

ln|2y+1| = x^2 + ln3 - 1
:. |2y+1| = e^(x^2 + ln3 - 1)

Noting that the exponential function is positive over its entire domain, (as e^x gt 0 AA x in RR):

2y+1 = e^(x^2 - 1)e^(ln3)

:. 2y = 3e^(x^2 - 1) - 1

:. y = 3/2e^(x^2 - 1) - 1/2