# How do I integrate dy/dx = x - y please ?

### 3 Answers | Add Yours

dy/dx = x - y

Use substitution:

v = x - y

Differentiate with respect to x

dv/dx = 1 - dy/dx

dy/dx = 1 - dv/dx

Now we use above substitutions in differential equations

dy/dx = x - y

1 - dv/dx = v

dv/dx = 1 - v

dv/(1-v) = dx

Now integrate both sides:

intg dv/(1-v) = intg dx

-ln (1-v) = x+c

ln(1-v) = -x - c

1 - v = e^(-x-c) = [e^(-x)][e^(-c)]

1 - v = (C)e^(-x) , where C = e^(-Cc)

1 - (x - y) = (C)e^(-x)

y - x + 1 = (C)e^(-x)

y = (C)e^(-x) + x - 1

Here is an example with a similar problem.

dy/dx = x-y is a differential equation. so you have to solve the differential equati

Solution :

We rewrite the equation as:

dy/dx+**1***y = x. So the integrating factor is e power Int dx = e^x..

Therefore ,

ye^x = Int (xe^x) dx = xe^x - Int (x)' e^x dx = xe^x- e^x +C . Or

y e^x = (x-1)e^x +C

dy/dx = x-y is a differential equation. so you have to solve the differential equati

Solution :

We rewrite the equation as:

dy/dx+**1***y = x. So the integrating factor is e power Int dx = e^x..

Therefore ,

ye^x = Int (xe^x) dx = xe^x - Int (x)' e^x dx = xe^x- e^x +C . Or

y e^x = (x-1)e^x +C