** Solving x'(t) = f(y(t))y'(t) = g(x(t)) **

Sometimes using algebra can be more efficient than using Laplace transforms in solving systems of differential equations.

Procedure - two paths we can take:

1. Take the derivative of x', x" = f'(y)

2. Solve for y' and equate it to g(x).

3. Solve for x(t).

4. Solve for y(t).

OR

1. Take the derivative of y', y" = g'(x)

2. Solve for x' and equate it to f(y).

3. Solve for y(t).

4. Solve for x(t).

Let's take some examples and see how this works. In all examples presented, x and y are functions of t.

1.

x'= 2y

y'= -2x

x" = 2y'

Solving for y':

y' = x"/2

Equate to g(x):

x"/2 = -2x

0 = x"/2 - 2x

0 = x" - 4x

We have a homogenous second order differential equation.

Characteristic Equation: λ^2 + 4 = 0

Roots: λ = ± 2i

Hence

x = C1 cos(2t) + C2 sin(2t)

Note x' = 2y.

y = x'/2

y = [d/dt C1 cos(2t) + C2 sin(2t) ] /2

y = -C1 sin(2t) + C2 cos(2t)

Solution:

x = C1 cos(2t) + C2 sin(2t)

y = -C1 sin(2t) + C2 cos(2t)

The practical approach to this type of problem is to end up working with a homogenous second order differential equation, if possible.

2.

x' = y + 1

y' = 2x

Note

x" = y'

Then

y' = x" = 2x

0 = x" - 2x

We have a homogenous second order equation where the characteristic equation is

λ^2 - 2 = 0 and λ = ±√2 (-√2 and √2).

So

x = C1 e^(√2 * t) + C2 * e^(-√2 * t)

With

x' = √2 * C1 * e^(√2 * t) - √2 * C2 * e^(-√2 * t) = y + 1

y = √2 * C1 * e^(√2 * t) - √2 * C2 * e^(-√2 * t) - 1

Solution:

x = C1 e^(√2 * t) + C2 * e^(-√2 * t)

y = √2 * C1 * e^(√2 * t) - √2 * C2 * e^(-√2 * t) - 1

If we run into a non-homogenous second order differential equation, its not the end other world, refer to Part 8 ( http://edspi31415.blogspot.com/2013/09/differential-equations-8-second-order.html) on how to deal with such equations.

The next example I purposely set it up so we are going to use variation of parameters.

3.

x' = y + 3

y' = -3x

Observe that y" = -3x' and x' = -y"/3, and

y"/3 + y + 3 = 0

y" + 3y + 9 = 0

y" + 3y = -9

y = y_h + y_p

y_h:

λ^2 + 3 = 0, λ = ± i*√3

y_h =C1 cos(t * √3) + C2 sin (t * √3)

y_p:

y1 = cos(t * √3)

y2 = sin(t * √3)

f(t) = -9

W = sin(t * √3) * d/dt cos(t * √3) - cos(t * √3) * d/dt sin(t * √3) = -√3

∫ y2 * f(t)/W dt = ∫ sin(t*√3) * -9 / -√3 dt = -3 * cos(t * √3)

∫ y1 * f(t)/W dt = ∫ cos(t*√3)*-9/-√3 dt = 3 * sin(t * √3)

y_p = cos(t * √3) * -3 * cos(t * √3) - sin(t * √3) * 3 * sin(t * √3) = -3

y = C1 cos(t * √3) + C2 sin (t * √3) - 3

We only have to do the variation of parameters once.

y' = -C1 * √3 * sin(t * √3) + C2 * √3 * cos(t * √3) = -3x

x = √3/3 * C1 * sin(t * √3) - √3/3 * C2 * cos(t * √3)

Solution:

x = √3/3 * C1 * sin(t * √3) - √3/3 * C2 * cos(t * √3)

y = C1 cos(t * √3) + C2 sin (t * √3) - 3

That concludes Part 9 of our series. Until next time,

Eddie

This blog is property of Edward Shore. 2013

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