Solving Simpler Differential Equations

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 (1 +  x 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 (1 +  x 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 = cx² - 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 (cx² – x) + x   (integrating)

2cx² – x = 2cx² -2x + x

And:

2cx² – x =  2cx² - x

Hence y = cx² – 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