__Preliminaries: __

A *differential **equation* is an equation that involves one or more derivatives or differentials. For example:

i)_ dy = x dx

is a differential equation. So is:

(ii) dQ/dt = - kQ

Also:

(iii) L(dI/dt) + RI = E

And: (iv) F = d(mv)/ dt

A function y = f(x) is a solution of a differential equation (DE) if the DE is identically satisfied when y and its derivatives are replaced throughout by f(x) and its corresponding derivatives. For example, in the case of (i) the solution is accomplished via the process called *integration*. If we integrate both sides, we obtain:

ò dy = ò x dx

Yielding:

y = x^{ 2} / 2 + c

where c is some undefined (as yet) constant of integration. We call the above the "general solution" to the differential equation. This general solution is, in fact, yields *a family of parabolas:*

**General solution - family of parabola - for dy = x dx.**

If we wanted to obtain the *particular* solution, we'd have to assign *boundary conditions* at the outset. Usually these designate what values x, y are to have at a particular point, and also often the first derivative (y' or dy/dx) at the same point.

We now look at another basic example:

Find the general solution of: dy/dx = 3 x^{ 2}

We rewrite in the form:

dy = 3 x^{ 2} dx

Then integrate:

ò dy = ò 3 x^{ 2} dx

y = x^{ 3} + c

Another example slightly more difficult:

Find the general solution of:

xy (1 + y^{ 2}) dx - (1 + x^{ 2}) dy = 0

Separate variables by dividing through by:

y (1 + y^{ 2})(1 + x^{ 2})

To obtain:

x dx/(1 + x^{ 2}) - dy/ y (1 + y^{ 2}) = 0

Integrate to get:

½ ln ^{ 2}) - ½ ln y^{ 2 }/1 + y^{ 2} = const.

This can be further simplified if we choose the constant to be: ½ ln c

Then multiply the resulting equation by 2 - and apply the properties of logarithms to get:

ln ^{ 2}) - ln y^{ 2 }/1 + y^{ 2} = ln c

Or:

ln (1 + x^{ 2}) (1 + y^{ 2})/ y^{ 2} = c

But since two numbers that have the same logarithms are equal:

(1 + x^{ 2}) (1 + y^{ 2})/ y^{ 2} = c

Or:

(1 + x^{ 2}) (1 + y^{ 2}) = c y^{ 2}

Lastly, we look at the following problem:

Show
that y = cx^{2} - x is a solution of the DE:

xy (dy/dx) = 2y + x

We first take (by implicit differentiation):

dy/dx = y’ = 2cx - 1

Then substitute for y and y’ into the DE:

xy' = 2y + x

So: x (2cx – 1) = 2 (cx^{2} – x)
+ x (integrating)

2cx^{2} – x = 2cx^{2} -2x + x

And:

2cx^{2} – x = 2cx^{2} - x

Hence
y = cx^{2} – x is a solution.

**Suggested Problems:**

Solve each of the following:

1) Ö(2 xy) (dy/dx) = 1

2) sin x (dx/dy) + cosh (2y) = 1

3) ln x (dx/dy) = x/y

4) dy/dx = exp (x) exp (-y)

5) Find the particular solution satisfied by the given condtions:

x dx + y dy = 0; y = 2 when x = 1

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