How do I solve a differential equation with unit step function?

this is an excellent DE for using Laplace transforms....if you do not know how, go to the library and get a book on these transforms and read it...if you do know how you can refresh yourself by using ' wikipedia.org'..of course if the domain of interest is (0,?) then your u(t) = 1

You can use Laplace transform methods to find the solution to this diff eq, but (and this can be a BIG but) the resulting solution is only valid for t > 0. If you need a solution for t< 0, it's not a big deal, but you need to resort to other solution methods. Indeed, for this problem, other methods are actually easier to use. Overall, then, I would disagree with Ted that this is an "excellent" problem upon which to use Laplace transform methods.

The equation to be solved is:

x'' +2x' + x = 10 * t * u(t)

where u(t) is the Heaviside step function.

If we don''t use Laplace transform methods, then the first step would be to solve the homogeneous problem, which, in this case, is also solution to the problem for t < 0:

x'' + 2x' + x = 0

Assuming a trial solution of the form x = exp(-k*t) yields the characteristic equation:

k^2 + 2k + 1 = 0

(k+1)^2 = 0

k= -1

The characteristic equation has a repeated real root, so the solution to the homogeneous problem, *and* the general solution for t<=0 is:

y(t) = A*exp(-t) + B*t*exp(-t)

For the case t<= 0, we can use the initial conditions to solve for the constants A and B:

y(0) = 1 = A + 0

A = 1

y'(t) = -A*exp(-t) + B*exp(-t) - B*t*exp(-t)

y'(0) = 0 = -A + B

B = 1

so for t<0, the solution is:

y(t) = exp(-t) + t*exp(-t); t<= 0

Now, for t> 0, the particular solution can easily be found using the method of undetermined coefficients. We form a trial particular solution consisting of a linear combination of the forcing term and all it's derivatives:

y_p(t) = a*t + b

Determine the values of the constants a and b by plugging this solution into the original differental equation and solving for them. To do this, we'll need the first and second derivatives of this trial solution:

y_p'(t) = a

y''_p(t) = 0

Now plug these into the nonhomogeneous equation describing the system for t>=0:

0 + 2a + at + b = 10t

Collecting like powers of t, we have:

at + (2a + b) = 10t

Equating the coefficients of like powers of t, gives

a = 10

2a+b = 0

So 20 + b = 0

b = -20

The general solution for t >0 is then:

y(t) = A*exp(-t) + B*t*exp(-t) + 10*t -20

Again, use the inital conditions to find A and B:


y(0) = 1 = A - 20

A = 21

y'(t) = -21*exp(-t) + B*exp(-t) - B*t*exp(-t) + 10

y'(0) = 0 = -21 + B + 10

0 = B-11

B = 11

So the general solution for t>=0 is:

y(t) = 21*exp(-t) + 11*t*exp(-t) + 10*t - 20; t >= 0


We can write a general solution for the entire domain of time as:


y(t) = exp(-t) + t*exp(-t) + u(t)*[20*exp(-t) + 10*t*exp(-t) + 10*t - 20]

y(t) = exp(-t) + t*exp(-t) + u(t)*[20*(exp(-t) - 1) + 10*t*(exp(-t) + 1]

Where, again, u(t) is the Heaviside step function.

You can crank through the Laplace transform method and will find that the solution you obtain is exactly the same as the solution obtained above for the nonhomogeneous case when t > 0. You'll need to use partial fractions to simplify the expression for the solution of the transformed equation, but then it's straightforward to use a table of transforms to find the inverse transform and the solution for t>0.