Another example where the nonhomogeneous part is a polynomial. Created by Sal Khan.
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- It might sound a little silly, but what happens when when you don't have 0 or f(t) on the right side, but a constant c? Do you treat c like an undetermined coeeficient, which would be then y=A, y'=1, y''=0 => A+1=c => A=1-c, if the D.E. was y''+y'+y-c=0?(14 votes)
- If you have just a constant on the right side, you can just move it to the other side and include it in your homogeneous portion.(12 votes)
- Did you half-drop a negative in substituting back in the original equation for y prime? I believe it should be +6Ax-3B(7 votes)
- No, I think it is correct as is. The "=" in the y prime equation just kind of looks like subtraction in the video. So substituting 2Ax + B into the -3y' part gives -6Ax - 3B.(11 votes)
- What if the right hand side is just x or 4x? Is it Ax+B then for the particular?(4 votes)
- m so confused how would i guess for particular sol.?(2 votes)
- For a particular solution, you guess with the same function given on the Right Hand Side with an unknown coefficient in front.
For example, as in this one, Sal uses the base function of x^2, which is really x^2 + 0x + 0 and puts an unknown coefficient in front of each variable. Therefore, Ax^2 + Bx + C is his guess.
In past examples, if the Right Hand Side was 4e^(3x), we would guess Ae^(3x) If the Right Hand Side was sin(5x), we would guess Asin(5x) + B cos(5x)...Hope this helps!(5 votes)
- What would I do if I have to solve for Y(0)= 1.6 and Y'(0)=4. Do I apply this to the particular solution, or to the general for the homogenous or both?(2 votes)
- You would apply this to both. ( in this video, Sal starts writing "y=" at6:15. This is "the solution") You have the "y=" as your first equation, then take the derivative of that "y'=" for your second equation.
All that is left is to plug 'n' chug.
Set the first equation "y=" to 1.6, and set the second equation "y'=" to 4.
Plug in 0 for your independent variable in "y=" equation and also the "y'=" equation.
You know have 2 equations and two unknown variables, Solve for one, plug into the other equation, solve for the second, Viola!(3 votes)
- Does anyone know what skill level these problems are? Would these be basic first year uni questions?(1 vote)
- yeah , m 101% sure these videos would help and they obviously have skill level.... m studying at UET Pakistan which is of course engineering uni ... and m watching these videos for basic things and i have to say more for logics ..... bcz our maths Sir does not bother to tell us logic behind question and we are sitting like puppets ,(3 votes)
- Is it compulsory to add the general solution of homogeneous eqn in the final answer? Because the examples in the previous videos, doesn't have one.(1 vote)
- Yes, the most general solution is always the sum of the homogenous solution plus the particular solution. If you limit the answer to only the particular solution, then you are ignoring a whole family of functions that are also solutions, and when the times comes to evaluate initial conditions, you won't be able to satisfy them.(2 votes)
- Hey, what should I use as my guess for solving the equation equal to xe^2x?(1 vote)
- If the non-homogenous differential equation is equal to x*e^(2x), I would start by guessing A*x*e^(2x). Of course, if c*x*e^(2x) is already a term in the homogenous solution, then you should guess A*x^2*e^(2x).(1 vote)
Let's do another example of solving a nonhomogeneous linear differential equation with a constant coefficient. And the left-hand side is going to be the same one that we've been doing. The second derivative of y minus 3 times the first derivative minus 4 times y is equal to-- and now instead of having an exponential function or a trigonometric functional, we'll just have a simple-- well, it just looks an x squared term, but it's a polynomial. Right? And you know how to solve the general solution of the homogeneous equation if this were 0. So we're going to focus just now on the particular solution, then we can later add that to the general solution of a nonhomogeneous equation, to get the solution. So what's a good guess for a particular solution? Well, when we had exponentials, we guessed that our solution would be an exponential. When we had trigonometric functions, we guessed that our solution would be trigonomretric. So since we have a polynomial here that makes this differential equation nonhomogeneous, let's guess that a particular solution is a polynomial. And that makes sense. If you take a second-degree polynomial, take its derivatives and add and subtract, you should hopefully get another second-degree polynomial. So let's guess that it is Ax squared plus Bx plus C. And what would be a second derivative? Well a second derivative would be 2Ax plus B. Sorry, this is the first derivitive. The second derivative would be 2A. And now we can substitute back into the original equation. We get the second derivitive, 2A minus 3 times the first derivitive. So minus 3A-- oh no, sorry. Minus 3 times this. So minus 6Ax minus 3B minus 4 times the function itself. So minus 4Ax squared minus 4Bx minus 4C. That's just 4 times all of that. That's going to equal 4x squared. And I'll just group our x squared, our x and our constant terms, and then we could try to solve for the coeficients. So let's see. I have one x squared term here. So it's minus 4Ax squared. And then what are my x terms? I have minus 6Ax minus 4Bx. So then say plus minus 6A minus 4B times x. I just added the coefficients. And then finally we get our constant terms. 2A minus 3B minus 4C. And all of that will equal 4x squared. Now how do we solve for A, B, and C? Well, whatever the x squared coefficients add up on this side, it should equal 4. Whatever the x coefficients add up on this side, it should be equal to 0, right? Because you can view this as plus 0x, right? And then you could say plus 0 constant as well. So the constants should also add up to 0. So let's do that. So first let's do the x squared term. So minus 4A should be equal to 4. And then that tells us that A is equal to minus 1. Fair enough. Now the x terms. Minus 6A, minus 4B, that should be equal to 0. Right? So let's write that down. We know what A is, so let's substitute. So minus 6 times A, so minus 6 times minus 1. So that's 6 minus 4B is equal to 0. So we get 4B-- I'm just putting 4B on this side and then switching. 4B is equal to 6. And B is equal to-- 6 divided by 4 is 3/2. And then finally the constant term should also equal 0, so let's solve for those. 2 times A, that's minus 2. Minus 3 times B. Well, that's minus 3 times this. So minus 9/2 minus 4C is equal to 0. So let's see. I don't want to make a careless mistake. So this is minus 4 minus 9/2, right? That's minus 4/2 minus 9/2-- and we could take the 4C and put it on that side-- it's equal to 4C. What's minus 4 minus 9? That's minus 13/2. Minus 13/2 is equal to 4C. 4C, divide both sides by 4, and then you get C is equal to minus 13/8. And I think I haven't made a careless mistake. So if I haven't, then our particular solution, we now know. Well, let me write 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 Ax squared, so that's minus 1x squared. Ax squared plus Bx plus 3/2x plus C minus 13/8. So this is the particular solution. We solved for A, B, and C. We determined the undetermined coefficient. And now if we want the general solution, we add to that the general solution of the homogeneous equation. What was that? y prime minus 3y prime minus 4y is equal to 0. And we've solved this multiple times. We know that the general solution to the homogeneous equation is C1e to the 4x plus C2e to the minus x, right? You just take the characteristic equation r squared minus 3r minus 4. What did you get? You get r minus 4 times r plus 1, and then that's how you get minus 1 and 4. Anyway. So if this is the general solution to the homogeneous equation, this a particular solution to the nonhomogeneous equation. The general solution to the nonhomogeneous equation is going to be the sum of the two. So let's add that. So plus C1e to the 4x plus C2e to the minus x. So there you. 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 nonhomogeneous differential equation with constant coefficients. See you in the next video.