Back to Differential Equations

First, let's look at the odd problem solutions:

1)Show that y = cx^2 - x is a solution of the DE: xy' = 2y + x

We first take:

dy/dx = y’ = 2cx - 1

Then substitute for y and y’ into the DE:

xy' = 2y + x

-> x (2cx – 1) = 2 (cx^2 – x) + x

2cx^2 – x = 2cx^2 -2x + x

And:

2cx^2 – x = 2cx^2 - x
Hence y = cx^2 – x is a solution.


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

We re-arrange to obtain: -4y dy = xdx

Integrate to get: -2y^2 = x^2/2

Or: x^2/2 + 2y^2 = c

From this one can put in a set of different values of c to generate the family of curves appropriate to the equation.

That is shown in Fig. 1.

On inspection we see one curve goes through the point (1,1) and that is associated with the equation:

x^2/2 + 2y^2 = 2

Or:

x^2 + 4y^2 = 4


5) Find the particular solution for the equation:exp(x) sec(y)dx + (1 + e^x) sec (y) tan(y) dy = 0 (y = 60 deg when x = 3 )

We can simplify and re-arrange the equation to give:

exp(x) dx/ (1 + exp(x)) = - tan (y) dy

then integrate both sides :

ln(1 + e^x) + ln sec(y) = c

and :

c = (1 + e^x)/ cos (y)

Since : (1 + e^x) = c cos(y)

The particular solution is found :

(1 + e^x) = 2(1 + e^3) cos (y)


Substitute the values for x, y :

c = (1 + e^3)/ cos(60) = (1 + e^3)/ ½

c = 2 (1 + e^3)

Let’s now return to differential equations and recap from the perspective of what we’ve already seen:

Given a differential equation F(x,y, y’, y’’…..y^n)= 0

1) A solution containing n arbitrary constants is a “general solution” (the number of such constants equals the number of the highest derivative)
2) A solution obtained from specific values of the constant(s) is called a “particular solution”
3) A solution to the DE which is not obtainable from the general solution is called a singular solution.

From these definitions (especially 1) one can always “work backward” and obtain the original differential equation from a general solution with arbitrary constants c1, c2, etc. known as a “primitive”

For example, given a primitive:

y = c1e^x + c2 e^-x + 3x

find the original DE.

Since we know the general solution we can differentiate twice to obtain the actual DE;

Differentiate once to get:

y’= c1e^x – c2e^-x + 3

and again:

y’’= c1e^x + c2e^-x

Now take:

y’’- y = c1e^x + c2e^-x - (c1e^x + c2 e^-x + 3x)

so: y’’- y = 3x

or d^2x/dx^2 – y = 3x

is the differential equation.

This also shows the (general) basis to solve differential equations of higher order, by successive differentiations. This leads us now to discriminate between the order and the degree of a DE:

The order of a differential equation is the same as the highest ordered derivative in the equation. Thus, in the example above, the order is 2 (since we have y’’)

The degree of a differential equation is the algebraic degree of the highest ordered derivative. Thus, the example above shows an equation of the 2nd order and first degree.

Given the preceding, let's now move on to solving initial value problems - which also make use of the general solution, but now with some inital values imposed.

Example: Given that the general solution of:

dy/dx + y = 2x exp(-x) can be written:

y = (x^2 + c) exp(-x)

solve the inital value problem for y(0) = 2

This means, if y = (x^2 + c) exp(-x)

Now: x = 0 when y = 2, so:

2 = (0 + c) exp(0) = c

Then: y = (x^2 + 2) exp(-x)

then: y' = -(x^2 + c)exp(-x) + 2xexp(-x)



Example (2) : dy/dx = x^2 sin(y)

and y(-1) = -2


As before, we first re-arrange to obtain:

dy/ sin(y) = x^2 dx

which can be integrated (both sides) to yield:

ln(tan (y/2)) = x^3/ 3 + c

or ln(tan(y/2)) - x^3/3 = c

Now, x = -1 when y = -2 so:

ln(tan(-1)) - (-1)^3/3 = c

or:

ln (-pi/2) + 1/3 = c

or:

0.443 + pi(i) + 1/3 = c

so the solution is: ln(tan (y/2)) = x^3/ 3 + 0.443 + pi(i) + 1/3

or: ln(tan(y/2)) - x^3/3 = 0.776 + pi(i)

Implicit differentiation often also finds its way into solving simple first order differential equations.

Consider, for example: dy/dx = -1/2y

One way is to re-arrange to get: 2y dy = -dx

and integrate both sides to get: y^2 = - x + c

Or: y^2 + x = c

On the other hand one may also use implicit differentiation (which will come in handy for more complex problems):

2y(x) y'(x) = -1

and integrate:

2y(x) y'(x)dx -> y^2(x) + x = c -> y^2 + x = c

Example: Show that x^3 + 3xy^2 = 1 is an implicit solution of the differential equation:
(2xy)dy/dx + x^2 + y^2 = 0

on the interval 0 less than x less than 1
We see why that interval is given by sketching the curve (Fig. 3). It is only well-behaved (continuous) between x slightly larger than 0 and slightly less than 1.

We begin by differentiating implictly to obtain:

3x^2 + 3y^2 + 6xy (dy/dx) = 0

divide through by 3: x^2 + y^2 + 2xy(dy/dx) = 0

Then:

2xy(dy/dx) + x^2 + y^2 = x^2 + y^2 + 2xy(dy/dx)


A few problems:

(1) Find the differential equation which has:

y = c1e^2x + c2e^-3x + sinx as the general solution

(2) State the degree and order of the differential equation:

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

and find the general solution.

(3) Show that 5x^2y^2 - 2x^3 y^2 = 1 is an implicit solution of the differential equation:

x (dy/dx) + y = x^3 y^3

over the interval (0, 5/2)

(also sketch the graph)

Solutions next time!