Let’s do another example of solving a non-homogeneous linear differential equation with constant coefficients and the left hand sides will be going to be the same one that we’ve been doing.
The second derivative Y minus three times first derivative minus four times the Y is equal to― and now instead of having a an exponential function or trigonometric function, we’ll just have a simple― well this is― it’s just look good X² term but it’s a polynomial, right. And you know how to solve the general solution of the homogeneous equation if this were zero. So we’re going to focus just now on the particular solution and we can later add that to the general solution of homogeneous equation to get the solution.
So let’s― what’s a good guess for a particular solution? Well, when we have exponentials, we guessed that our solution would be an exponential and we have trigonometric functions. We guess that our solution would be trigonometric. So since we have a polynomial here that makes this differential equation non-homogeneous, let’s guess that a particular solution is a polynomial and that makes sense if you take a polynomial of and actually a second degree polynomial.
If you take a second degree polynomial, take its derivatives, add and subtract you should hopefully get another second degree polynomials. So let’s guess that it is AX² plus BX plus C and what would be a second derivative, well second derivative would be two AX plus B and then the first derivative, the second derivative would be two A. And now we could substitute back into the original equation, we get the second derivative, two A minus three times the first derivative so minus three times this, so minus six AX minus three B, minus four times the function itself so minus four AX² minus four BX minus four C.
That’s just four times all that, that’s going to equal four X², now let’s group our X², our X and our constant term and then we could try to solve for the coefficients. So let’s see, I have one X² term here, so it’s minus four AX² and then what are my X terms? I have minus six AX, minus four BX so that’s say plus minus six A minus four B times X, I just added the coefficients. And then finally, we get our constant terms, two A minus three B minus four C. So plus two A minus three B minus four C and all of that will equal four X².
Now how do we solve for AB and C? Well whatever the X² coefficients add up on this side, it should equal four, whatever the X coefficient adds up on this side, it should be equal to, well it should be equal to zero, right. Because you can view this as plus zero X, right. And then you could say plus zero constant as well. So the constant should also add up to zero. So let’s do that, so first let’s do the X² term. So minus four A should be equal to four, so minus four A is equal to four and then that tells us that A is equal to minus one, fair enough. Now, the X terms this minus six A minus four B that should be equal to zero, right.
So, let’s write that down, we know what A is so let’s substitute. So minus six times A, so minus six times minus one, so that’s six minus four B is equal to zero. So we get four B, I’m just putting four B on this side and then switching, four B is equal to six and B is equal to six divided by four is three over two. And finally, the constant term should also equal zero. So let’s add those, let’s solve for those, well two times A, that’s minus two, minus three times B, well that’s minus three times this. So minus nine halves minus four C is equal to zero so let’s see.
I don’t want to make a careless mistake. So this is minus four, minus nine over two, right that’s minus four over two minus nine over two and then we could take the four― slip it on that side is equal to four C and that’s what’s minus four minus nine, that’s minus 13 over two. So minus 13, over two is equal to four C or C divide both sides by four and then you get C is equal to minus 13 over eight. And I think I haven’t made a careless mistake so I haven’t then our particular solution we now know. And actually let’s just right it in; let me write them, the whole solution. So and this is a nice stretch of horizontal real estate, so let’s write our solution.
Our solution is going to be equal to the particular solution, which is AX². So that’s minus one X², right AX² plus BX plus three halves X plus C minus 13 over eight. So this is a particular solution, we solve for AB and C, we determined the undetermined coefficient and now if you want the general solution, we add to that the general solution of the homogeneous equation. The general solution of the homogeneous equation, what was that Y prime minus three Y prime minus four Y is equal to zero and we’ve solve this multiple times, we know that the general solution, the homogeneous equation is C one E to the four X plus C two E to the minus X, right.
You just take the characters equation R² minus three R minus four you get what you get, you get R minus four times R plus one and then that’s how you get minus one and four. Anyway, so this, the general solution to the homogeneous equation, this is A particular solution to the non-homogeneous equation. The general solution to the homogeneous equations would be the sum of the two so let’s add that so plus C one E to the four X plus C two E to the minus X. So there you go.
I don’t think that was too painful, the most painful part was just making sure that you don’t make a careless mistake with the algebra but using a fairly straightforward, really algebraic technique we were able to get a fairly fancy solution to this second order linear non-homogeneous differential equation with the constant coefficients. See you in the next video.