Differential Equations

Example 3

Solve: dy + 2ydx = 6 dx given x = 0 when y = 1.

Separate variables:

dy = (6 – 2y) dx

dy/(6 – 2y) = dx

Integrate, solve for y, then C:

-ln (6 – 2y)/2 = x + C

ln (6 – 2y) = 2x + 2C

Note that 2C is still a constant. Let’s call it C’.

ln (6 – 2y) = 2x + C’

Now take the exponential of both sides:

6 – 2y = e^{-2x + C’}

6 – 2y = e^{-2x} * e^{C’}

And e^{C’} is a constant. Let’s call it C”.

6 – 2y = e^{-2x} * C”

Or: 6 – 2y = C”e^{-2x}

This is a good place to solve for our constant, C”.

6 – 2(1) = C”e^{-2(0)}

4 = C”

6 – 2y = 4e^{-2x}

Solve for y: y = 3 – 2e^{2x}

Whew! That was a long one. Study it over again. We’ll be seeing more like it later.