Consider a differential equation of type
\[y^{\prime\prime} + py' + qy = 0,\]
where p, q are some constant coefficients.
For each of the equation we can write the so-called characteristic (auxiliary) equation:
\[{k^2} + pk + q = 0.\]
The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are the following options:
- Discriminant of the characteristic
quadratic equation \(D \gt 0.\) Then the roots of the characteristic equations \({k_1}\) and \({k_2}\) are real and distinct. In this case the general solution is given by the following function
\[y\left( x \right) = {C_1}{e^{{k_1}x}} + {C_2}{e^{{k_2}x}},\]
where \({C_1}\) and \({C_2}\) are arbitrary real numbers. - Discriminant of the characteristic quadratic equation \(D = 0.\) Then the roots are real and equal. It is said in this case that there exists one repeated root \({k_1}\)
of order 2. The general solution of the differential equation has the form:
\[y\left( x \right) = \left( {{C_1}x + {C_2}} \right){e^{{k_1}x}}.\]
- Discriminant of the characteristic quadratic equation \(D \lt 0.\) Such an equation has complex roots \({k_1} = \alpha + \beta i,\) \({k_2} = \alpha - \beta i.\) The general solution is written as
\[y\left( x \right) = {e^{\alpha x}}\left[ {{C_1}\cos \left( {\beta x} \right) + {C_2}\sin \left( {\beta x} \right)} \right].\]
Solved Problems
Click or tap a problem to see the solution.
Example 1
Solve the differential equation \[y^{\prime\prime} - 6y' + 5y = 0.\]
Example 2
Find the general solution of the equation \[y^{\prime\prime} - 6y' + 9y = 0.\]
Example 1.
Solve the differential equation \[y^{\prime\prime} - 6y' + 5y = 0.\]
Solution.
First we write the corresponding characteristic equation for the given differential equation:
\[{k^2} - 6k + 5 = 0.\]
Eliminate the constant \(C\) from the system of equations:
\[y\left( x \right) = {C_1}{e^x} + {C_2}{e^{5x}},\]
where \({C_1}\) and \({C_2}\) are arbitrary constants.
Example 2.
Find the general solution of the equation \[y^{\prime\prime} - 6y' + 9y = 0.\]
Solution.
We write the characteristic equation and calculate its roots:
\[{k^2} - 6k + 9 = 0,\;\; \Rightarrow D = 36 - 4 \cdot 9 = 0,\;\; \Rightarrow {k_{1,2}} = 3.\]
As it can be seen, the characteristic equation has one root of order \(2:\) \({k_1} = 3.\) Therefore, the general solution of the differential equation is given by
\[y\left( x \right) = \left( {{C_1}x + {C_2}} \right){e^{3x}},\]
where \({C_1},\) \({C_2}\) are arbitrary real numbers.
See more problems on Page 2.
Video transcript
We'll now move from the world of first order differential equations to the world of second order differential equations. And what does that mean? That means that we're now going to start involving the second derivative. And the first class that I'm going to show you-- and this is probably the most useful class when you're studying classical physics-- are linear second order differential equations. So what is a linear second order differential equation? I think I touched on it a little bit in our very first intro video. But it's something that looks like this. If I have a of x-- so some function only of x-- times the second derivative of y, with respect to x, plus b of x, times the first derivative of y, with respect to x, plus c of x, times y is equal to some function that's only a function of x. So just to review our terminology, y is the second order because the highest derivative here is the second derivative, so that makes it second order. And what makes it linear? Well all of the coefficients on-- and I want to be careful with the term coefficients, because traditionally we view coefficients as always being constants-- but here we have functions of x as coefficients. So in order for this to be a linear differential equation, a of x, b of x, c of x and d of x, they all have to be functions only of x, as I've drawn it here. And now, before we start trying to solve this generally, we'll do a special case of this, where a, b, c are constants and d is 0. So what will that look like? So I can just rewrite that as A-- so now A is not a function anymore, it's just a number-- A times the second derivative of y, with respect to x, plus B times the first derivative, plus C times y. And instead of having just a fourth constant, instead of d of x, I'm just going to set that equal to 0. And by setting this equal to 0, I have now introduced you to the other form of homogeneous differential equation. And this one is called homogeneous. And I haven't made the connection yet on how these second order differential equations are related to the first order ones that I just introduced-- to these other homogeneous differential equations I introduced you to. I think they just happen to have the same name, even though they're not that related. So the reason why this one is called homogeneous is because you have it equal to 0. So this is what makes it homogeneous. And actually, I do see more of a connection between this type of equation and milk where all the fat is spread out, because if you think about it, the solution for all homogeneous equations, when you kind of solve the equation, they always equal 0. So they're homogenized, I guess is the best way that I can draw any kind of parallel. So we could call this a second order linear because A, B, and C definitely are functions just of-- well, they're not even functions of x or y, they're just constants. So second order linear homogeneous-- because they equal 0-- differential equations. And I think you'll see that these, in some ways, are the most fun differential equations to solve. And actually, often the most useful because in a lot of the applications of classical mechanics, this is all you need to solve. But they're the most fun to solve because they all boil down to Algebra II problems. And I'll touch on that in a second. But let's just think about this a little bit. Think about what the properties of these solutions might be. Let me just throw out something. Let's say that g of x is a solution. So that means that A times g prime prime, plus B times g prime, plus C times g is equal to 0. Right? These mean the same thing. Now, my question to you is, what if I have some constant times g? Is that still a solution? So my question is, let's say some constant c1 gx-- c1 times g-- is this a solution? Well, let's try it out. Let's substitute this into our original equation. So A times the second derivative of this would just be-- and I'll switch colors here; let me switch to brown-- so A times the second derivative of this would be-- the constant, every time you take a derivative, the constant just carries over-- so that'll just be A times c1 g prime prime, plus-- the same thing for the first derivative-- B times c1 g prime, plus C-- and this C is different than the c1 c-- times g. And let's see whether this is equal to 0. So we could factor out that c1 constant, and we get c1 times Ag prime prime, plus Bg prime, plus Cg. And lo and behold, we already know. Because we know that g of x is a solution, we know that this is true. So this is going to be equal to 0. Because g is a solution. So if this is 0, c1 times 0 is going to be equal to 0. So this expression up here is also equal to 0. Or another way to view it is that if g is a solution to this second order linear homogeneous differential equation, then some constant times g is also a solution. So this is also a solution to the differential equation. And then the next property I want to show you-- and this is all going someplace, don't worry. The next question I want to ask you is, OK, we know that g of x is a solution to the differential equation. What if I were to also tell you that h of x is also a solution? So my question to you is, is g of x plus h of x a solution? If you add these two functions that are both solutions, if you add them together, is that still a solution of our original differential equation? Well, let's substitute this whole thing into our original differential equation, right? So we'll have A times the second derivative of this thing. Well, that's straightforward enough. That's just g prime prime, plus h prime prime, plus B times-- the first derivative of this thing-- g prime plus h prime, plus C times-- this function-- g plus h. And now what can we do? Let's distribute all of these constants. We get A times g prime prime, plus A times h prime prime, plus B times the first derivative of g, plus B times the first derivative of h, plus C times g, plus C times h. And now we can rearrange them. And we get A-- let's take this one; let's take all the g terms-- A times the second derivative of g, plus B times the first derivative, plus C times g-- that's these three terms-- plus A times the second derivative of h, plus B times the first derivative, plus C times h. And now we know that both g and h are solutions of the original differential equation. So by definition, if g is a solution of the original differential equation, and this was the left-hand side of that differential equation, this is going to be equal to 0, and so is this going to be equal to 0. So we've shown that this whole expression is equal to 0. So if g is a solution of the differential equation-- of this second order linear homogeneous differential equation-- and h is also a solution, then if you were to add them together, the sum of them is also a solution. So in general, if we show that g is a solution and h is a solution, you can add them. And we showed before that any constant times them is also a solution. So you could also say that some constant times g of x plus some constant times h of x is also a solution. And maybe the constant in one of the cases is 0 or something. I don't know. But anyway, these are useful properties to maybe internalize for second order homogeneous linear differential equations. And in the next video, we're actually going to apply these properties to figure out the solutions for these. And you'll see that they're actually straightforward. I would say a lot easier than what we did in the previous first order homogeneous difference equations, or the exact equations. This is much, much easier. I'll see you in the next video.