Math Revisited: Looking At Differential Equations

Let's consider one of the simplest differential equations:

xdx + ydy = 0

One could be understandably tempted to write this in terms of the 1st derivative to obtain:

dy/dx = -x/y

but this serves no useful purpose. It's more productive to directly integrate the equation: xdx = -ydy, viz.

ò x dx   =   ò-y dy

(since variables are already separated) to obtain:

x/2 = - y² / 2 ,  so

x² + y² = c

where c is the constant of integration.

Now, say the particular conditions are such that y = Ö2 when x = Ö2  then:

(Ö2)² + (Ö2)² = 2 + 2 = 4

so the particular solution is: x² + y² = 4

Which is the equation for the circle, e.g.

Now let's examine a more complex 1st order equation :

Find the general solution and the particular curve passing through the point (0,0) of the differential equation:

exp(x) cos(y) + (1 + exp(x)) sin(y) dy = 0

This looks a bit fearsome, but again, the first rule is simplify, which means separating variables (this is also usually where one's acumen with basic algebra comes in!)

we obtain:

exp(x)/ (1 + exp(x)) + [sin(y)/ cos(y)] dy

= exp(x)/ (1 + exp(x)) + tan(y)dy = 0


We then integrate:  ò (e^x)/ (1 + e^x) dx   =   ò tan (y) dy

To obtain:

ln(1 + e^x) - ln cos(y) = ln c

Or:  ln(1 + e^x) = ln c + ln cos(y)

where again, c is the constant of integration. We can easily simplify the above (using well known principles of natural logs) to get:

1 + e^x = c cos (y)

Which is the general solution.



To get the particular solution we need to substitute the ordered pair values for (0,0) into the general solution, whence:

1 + e⁰ = c(cos (0))

so that: 1 + 1 = c

and c = 2 ,    then we get: 1 + e^x = 2 cos (y)

Problems:

1)Show that y = cx² - x is a solution of the DE:

xy' = 2y + x

2) Show that y = (2x + c) exp(-x) is a solution of the DE: dy/dx + y = 2 exp(-x)

3) For the differential equation: dy/dx = -x/4y , sketch the curve which passes through the point (1,1)

4) Find the general solution of: sin (t) dp + p cos(t) dt = 0

5) Find the particular solution for the equation: 

exp(x) sec(y)dx + (1 + e^x) sec (y) tan(y) dy = 0

with  y = 60 deg when x = 3