Second Order Differential equations

i have never been taught how to do second order differential equations before and we have to teach it to ourselves (not going over it in lectures). i have tried looking on the internet for methods of solving but cant work out what its going on about :s-smilie:

can someone explain, using this random example i've thought of?

3y" + 5y' - 2y = 2x + 3

thanks :smile:

Reply 1

james.h

wrooru

can someone explain, using this random example i've thought of?

3y" + 5y' - 2y = 2x + 3

thanks :smile:

Okay, for any 2nd order ODE (ordinary differential equation), you'll want to start off by finding the related auxiliary equation. This auxiliary equation is basically just a quadratic describing the ODE, replacing the terms y'', y' and y with m^2, m and 1. So for this example, the auxiliary equation is:

\displaystyle 3m^2 + 5m - 2 = 0

(Note that, for the time being, we're just considering the right-hand side of the ODE; the left-hand side will be considered afterwards.)

Now, all you need to do is solve (factorise) that quadratic:

\displaystyle (m+2)(3m-1) = 0

Hence you get

m_1=-2

and

m_2=\frac{1}{3}

as your solutions. These solutions are combined to give what is termed the particular integral, as follows:

\displaystyle y_1 = Ae^{m_1x} + Be^{m_2x} = Ae^{-2x} + Be^{\frac{1}{3}x}

, where A and B are constants.

Now, if the left-hand side is zero, you just stop here and the general solution is the particular integral as you have just found (above). However, in this example there is a function on the LHS as well, so we proceed to find what is called the complementary function to take this extra part of the ODE into account...

First, try a solution of the form

y_2 = Cx + D

(a polynomial of degree 1, like the left-hand side). You can now differentiate this twice to obtain expressions for both

y_2'

and

y_2''

. Substituting these values into the original ODE (from the question), gives:

\displaystyle 3(0) + 5(C) - 2(Cx + D) = 2x + 3

You can now solve to find the values of the constants

C

and

D

. With that done, it's a simple matter of applying the principle of superposition to get the final answer; the general solution will be equal to the particular integral plus the complementary function, like so:

\displaystyle y = y_1 + y_2 = Ae^{-2x} + Be^{\frac{1}{3}x} + Cx + D

, where you've found the values of

C

and

D

(I can't be bothered to do this at the moment :p:

).

If you are given initial conditions (for example, when y = 0, y' = 1 and y'' = 2) then you can apply these - again by susbtitution - to your solution to find the values of

A

and

B

. Otherwise, you're done. :biggrin:

Differential equations... :love:

PS: If you find the auxiliary equation has two equal roots, then the particular integral will be of the form:

Ae^{m_1x} + Bxe^{m_2x}

(note the extra factor x multiplying the second term).

PPS: If you're dealing with a trigonometric function (eg:

\sin(x)

on the LHS (for the complementary function), remember to try a solution of the form

C\sin(x) + D\cos(x)

, even though there is no cosine function on the LHS. This is because the derivatives of sine and cosine interchange with each other...if that makes sense. You probably know what I mean. :o: