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# Second Order Differential equations watch

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

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

thanks
2. (Original post by wrooru)
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 can someone explain, using this random example i've thought of?

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

thanks
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:

(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:

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

, 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 (a polynomial of degree 1, like the left-hand side). You can now differentiate this twice to obtain expressions for both and . Substituting these values into the original ODE (from the question), gives:

You can now solve to find the values of the constants and . 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:

, where you've found the values of and (I can't be bothered to do this at the moment ).

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 and . Otherwise, you're done.

Differential equations...

PS: If you find the auxiliary equation has two equal roots, then the particular integral will be of the form: (note the extra factor x multiplying the second term).

PPS: If you're dealing with a trigonometric function (eg: on the LHS (for the complementary function), remember to try a solution of the form , 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.

If you need any more help or details on the topic, just quote me.

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