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.

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 = - y

x

where c is the constant of integration.

Now, say the particular conditions are such that y =

(

so the

x/2 = - y

^{2}/ 2 , sox

^{2}+ y^{2}= cwhere c is the constant of integration.

Now, say the particular conditions are such that y =

**Ö**2 when x =**Ö**2 then:(

**Ö**2)^{2}+ (**Ö**2)^{2}= 2 + 2 = 4so the

*particular solution*is:**x**^{2}+ y^{2}= 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

We then integrate: ò (e

^{x})/ (1 + e^{x}) dx = ò tan (y) dy
To
obtain:

ln(1 + e

Or: ln(1 + e

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

Which is

To get the

1 + e

so that: 1 + 1 = c

and c = 2 , then we get: 1 + e

ln(1 + e

^{x}) - ln cos(y) = ln cOr: 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 + e

^{0 }= 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 = cx

^{2}- 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:

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

with y = 60 deg when x = 3

^{x}) sec (y) tan(y) dy = 0with y = 60 deg when x = 3

_{}

^{}

## No comments:

Post a Comment