Differential Equations (III)

We left off with several problems. Let’s look at the solutions:

(1) Find the differential equation which has: y = c1e^2x + c2e^-3x + sin(x) as the general solution

Solution:

Differentiate twice:

y’ =dy/dx = 2c1e^2x -3c2e^- 3x + cos (x)

y” = d^y/dx^2 = 4c1e^2x + 9c2e^-3x - sin (x)


Set the determinants of the coefficients of c1 and c2 and the constant terms = 0 (refer back to blogs on matrices, and determinants)

(y - sin (x) e^2x e^-3x)
(y’ -cos(x) 2e^2x -3e^-3x) = 0
(y” + sin(x) 4e^2x 9e^-3x)

Simplify:

e^2y(e^-3x) *

(y – sin(x) 1 1)
(y’ – cos(x) 2 -3)= 0
(y” + sin(x) 4 9)


-> e^-x[-5y” -5y’ +30y -35sin(x) +5 cos(x)]= 0


Leading to the DE: y” + y’ – 6y + 7sin(x) –cos(x) = 0


(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.

The degree is 1, and the order is 1. Separating variables one gets:

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

Integrating:

½ ln (1 + x^2) - ½ ln y^2/ (1 + y^2) = c

Inspection of the equation shows that if we choose the constant c = ½ ln c
We can make the solution very easy since multiplication by 2 then simplifies it immensely:

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


We now just need to apply the properties of logarithms to obtain:

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

and further:

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

or: cy^2 = (1 + x^2)(1 + y^2)


(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)

The graph here is sketched in the accompanying diagram: the shape shows why the interval must be so rigidly confined.

We differentiate implicitly to get:

10xy^2 + 5x^2 2yy’ – 6x^2y^2 + 2x^3 2yy’ = 0

And:

2yy’ (5x^2 + 2x^3) = 6x^2y^2 – 10xy^2

Then:

y’ = y^2(6x^2 – 10x)/ 2y (5x^2 + 2x^3)

= y(6x^2 – 10x)/ 2(5x^2 + 2x^3)

But: y(6x^2 – 10x)/ 2(5x^2 + 2x^3) = x^2y^3 – y/x

= (x^3y^3 – y/ x = dy/dx

So: x(dy/dx) = x^3y^3 – y or x(dy/dx) + y = x^3 y^3



Intro to Partial derivatives- Exact Differential Equations:

Alas, we have to introduce a bit about partial differentiation here, to make sense of what we call exact differential equations.

Definition:

Given some differential equation: M(x,y)dx + N(x,y)dy = 0

If there exists a function f(x,y) such that:

@f/@x = M(x,y) and @f/@y = N(x,y)

Then the differential equation is said to be exact. (where @f/@x, @f/@x are the partial derivatives of f with respect to x and y, respectively)

A necessary and sufficient condition that the DE is exact is also that:

@M/@y = @N/@x


As an example, we want to find out if: dy/dx = (x + y)/ xy^2 is exact

Re-arrange to get: (x + y)dx – xy^2dy = 0

Then: M(x,y) = (x + y)

N(x, y) = (-xy^2)

Now, the partial derivative with respect to x is taken by differentiating with respect to x, and holding y constant. Similarly, the partial derivative with respect to y is taken by differentiating with respect to y, and holding x constant.

Thus: @M/@y = x and @N/@x = -2yx

So this DE is not exact since @N/@x is not equal to @M/@y


Example (2) Show that the DE:

2xydx + (1 + x^2) dy = 0

is exact and find the general solution

As before: M(x,y) = 2xy and N(x.y) = (1 + x^2)

Then: @M/@y = 2x and @N/@x = 2x

So yes, the DE is exact!

To obtain the general solution, let:

f(x,y) = INT_x (2xy dx + c(y) = x^2y + c(y)

where INT denotes integral and the subscript implies with respect to x

since @f/@y = N we have:

@/@y [ x^2y + c(y)] = x^2 + d/dy [c(y)] = 1 + x^2

(Since: @/@y [ x^2y + c(y)] = x^2 and @c(y)/@y = 0)

We see: d/dy [c(y)] = 1 and c(y) = y (e.g. INT dc(y) = c(y) = INT dy = y)

The function (or general solution) is then:

f(x,y) = x^2y + c(y) = x^2y + y or x^2y +y = c


Some problems:

(1) State whether the DE: dy/dx = (2x + y^2)/ -2xy is exact

(2) Show that the DE below is exact and find the solution

(3xy^4 + x)dx + (6x^2y^3 – 2y^2 + 7)dy = 0

(3) Consider the differential equation:

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

Determine whether it is an exact DE or not. If it is, find the general solution and then the particular for an initial condition such that :

y(1) = 3.

Solutions next time, and we look at integrating factors!