Math Is Fun Forum

eluru

Member

Registered: 2008-11-02

Posts: 7

solution of diff equation.

can u plz give me the steps involved in  taking the solution of differential equation.
y= c1 exp(kx)+c2 exp(-kx)  into the form  y= c1 sinkx+ c2 coskx.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: solution of diff equation.

We are starting with a 2nd order homogeneous differential equation with constant coefficients:

And we come up with a characteristic equation with root:

Giving the general solution:

And this is a solution for all C_1 and C_2.  So let's set C_1 = 1 and C_2 = 0, and then use Euler's equation.

Now we do the same with setting C_1 = 0 and C_2 = 1:

The last step is just trig identities.  Now it is easy to prove that if y_1 and y_2 are solutions to a differential equation, then y_1 + y_2 and k*y_1 are solutions as well for any k.  So i*y_1 - i*y_2 is a solution.  Thus, we have two solutions:

Now use the Wronskian to prove that these two solutions are in fact linearly independent so long as b is not 0 (why can't b be 0?).  Once you do that, we have a theorem that the most general solution to the differential equation is expressed as a linear combination of two linearly independent solutions.  Therefore, the most general form is:

Since C_1 and C_2 are constants, the 2 and -2 are both "absorbed" leaving:

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