Second Order ODEs
Contents
14. Second Order ODEs¶
Definition
In general a linear second order differential equation is of the form:
where \(a,\,b,\,c\) are arbitrary constants.
We can also conveniently write these problems using a linear operator:
With this definition, the problems may be written as:
\(\mathcal{L}(y)=0\) (homogeneous)
\(\mathcal{L}(y)=f(x)\) (inhomogeneous)
14.1. Homogeneous problems¶
Firstly, we consider the case of (14.1) when \(f(x)=0\), where we obtain the homogeneous problem:
14.1.1. The trial solution¶
One way to start with this problem is by using the trial solution (also sometimes called the ansatz):
where \(A_\lambda\) is constant. If we try this, then the homogeneous problem reduces to a polynomial in \(\lambda\):
and so by solving this quadratic in \(\lambda\), we can find solutions to (14.2). We call this quadratic the auxiliary equation and the nature of the solutions depends on the quantity:
which if we recall is called the discriminant.
Worked example
For the differential equation:
where we apply the trial solution \(y = A_\lambda\,e^{\lambda x}\), the auxiliary equation is:
The quadratic here has roots
so we have found two particular solutions of the given ODE:
where the constants \(A_1,\,A_2\) are to be found by applying initial or boundary conditions.
Some motivation for the ansatz
We could motivate the form of the trial solution by solving the problem for the special case where \(c=0\), which reduces to:
This problem can be solved by separation and the result is of the form \(y=A_1e^{-bx/a}+A_2\).
Another way is to begin by writing the system in matrix form \(\underline{y}^{\prime}=A\,\underline{y}\), where \(\underline{y}=[\begin{matrix}y& y^{\prime}\end{matrix}]^T\) :
The eigenvalues of the problem \(A\,\underline{y}=\lambda\,\underline{y}\) satisfy the quadratic \(a\lambda^2+b\lambda +c=0\) and therefore the solutions to the problem \(\underline{y}^{\prime}=k\,\underline{y}\) are of an exponential form.
14.1.2. Principle of linear superposition¶
The general second order homogeneous ODE given is linear, meaning that it has no products of terms involving the dependent variable \(y\). As a result of this fact, any linear combination of solutions will also be a solution:
We sometimes call this solution the complementary function.
Worked example
For the differential equation:
the solutions are found to be:
By superposition, we can combine these two solutions to obtain a more general solution:
where \(A_1,\, A_2\) are constants to be found.
You are encouraged to check this solution to confirm that it works!
14.1.3. Applying conditions¶
Since the results we obtain from solving the auxillary equation and then making a linear superposition of solutions have two arbitrary constants, it generates a family of solution curves. We can fix the solution to one of these curves by providing conditions for either \(y\) or its derivatives at two different \(x\) values - these often go by many different names such as initial coditions or boundary conditions. For practical problems, the conditions tell us about a particular state of the system.
For instance, applying Newton’s second law to the motion of a pendulum gives a second order differential equation for the angular displacement \(\theta\) from the downward vertical. Conditions for \(\theta(0)\) and \(\dot{\theta}(0)\) specify the initial displacement and the initial angular velocity. These types of condition, specified at time \(t=0\) are called initial conditions. For some other types of problem we may have conditions specified at two end-points of an interval. Such types of condition are called boundary conditions.
Practice questions
The solutions to the ODE:
is given by:
1. Find the particular solution that satisfies \(y(0)=2,\, y^{\prime}(0)=1\)
2. Find the particular solution that satisfies \(y(0)=3,\, y(1)=e^{-1/2}+2e\)
Solutions
Given the general solution to the ODEs, we can find the first derivative:
1. Substituting for the given conditions:
\( y(0)=1\Rightarrow A_1+A_2=2\)
\( y^{\prime}(0)=1 \Rightarrow -\frac{1}{2}A_1+A_2=1\)
Solving these two equations simultaneously gives \(A_1=\frac{2}{3}\), \(A_2=\frac{4}{3}\).
The solution satisfying the given conditions is therefore given by \(y(x)=\frac{2}{3}e^{-\frac{1}{2}x}+\frac{4}{3}e^x\)
2. Substituting for the given conditions:
\( y(0)=3\Rightarrow A_1+A_2=3\)
\( y(1)=e^{-1/2}+2e \Rightarrow A_1e^{-1/2}+A_2e=e^{-1/2}+2e\)
Solving these two equations simultaneously gives \(A_1=1\), \(A_2=2\).
The solution satisfying the given conditions is therefore given by \(y(x)=e^{-\frac{1}{2}x}+2e^x\)
14.1.4. Complex conjugate roots \(\Delta < 0\)¶
If the discriminant is negative, then the auxiliary equation will have complex conjugate solutions of the form
In this case we can write the general solution a bit differently. We make use of our knowledge of complex numbers (Euler’s theorem) to rewrite:
upon relabelling the constants \(A_1,\, A_2 \rightarrow B_1,\, B_2\). This is a much nicer form of the solution, because all parts of it are real, and so we expect that the coefficients \(B_1,\ B_2\) will be real as well. It also shows us clearly that the solution consists of an exponentially growing/decaying amplitude combined with wavy oscillations of frequency \(\omega\).
Worked example
Find the general solution of the problem
By inserting the trial solution, we find the auxiliary equation:
has roots \(\lambda=1\pm i\). Therefore the general form of the solution is:
14.1.5. Repeated roots \(\Delta = 0\)¶
If the discriminant is zero, then the auxiliary equation will only have one distinct solution \(e^{\lambda x}\), where
We will look for another solution of the form:
In this expression, \(f(x)\) could be any function, so this step is perfectly legitimate. But we have done it because it will allow us to reduce the order of the differential equation by making use of the known solution. We obtain:
The first term disappears since \(e^{\lambda x}\) is a solution, and so to satisfy the problem we require:
Using the result for the repeated root \(\lambda\) reduces this equation to \(f^{\prime\prime}=0\), which has a solution of the form:
We can choose \(A=1,\, B=0\), thus we have found another solution \(x e^{\lambda x}\) and we can form the general solution from a linear combination:
It is very important you recognise that the only reason \(x e^{\lambda x}\) has been found as a solution here is because \(\lambda\) is a repeated root, which makes the term involving \(2a\lambda+b\) drop out. If you tried the same technique using one of the solutions to a problem with distinct roots it would not give you a result of the form \(xe^{\lambda x}\) (in fact after some messy first order integration you would simply recover the solution involving the other root).
Practice Questions
For the problem:
1. Find the general solution and verify by substituting into the ODE that your solution works.
2. Find the particular solution that satisfies \(y(0)=1,\, y^{\prime}(0)=3\).
Solutions
1. The auxiliary equation is \(\lambda^2 - 4\lambda + 4 = 0\), with repeated root \(\lambda=2\).
Hence, the general solution is given by \(y=e^{2x}(A_1\,x + A_2)\), where \(A_1\) and \(A_2\) are constants.
2. \(y(0)=1\Rightarrow A_1=1\),
\(y^{\prime}(0)=3 \Rightarrow 2A_1+A_2=3\)
The solution satisfying the given conditions is \(y-e^{2x}(x+1)\)
14.1.6. Existence of a particular solution:¶
The general solution:
can be made to satisfy arbitrary initial conditions \(y(x_0)=y_0,\ y^{\prime}(x_0)=y^{\prime}_0\) if and only if:
This condition is met if \(y_2 \neq \alpha y_1\), where \(\alpha\) would be a constant - we say that these solutions are linearly independent.
To prove that this is true, start by applying the given conditions:
This is a set of two simultaneous equations in two unknowns \(A_1,\, A_2\) - a unique solution exists if and only if the condition (14.4) is met.
Summary
The general solution of the homogeneous second order ODE is given by:
where \(\Delta=b^2-4ac\).
14.2. Inhomogeneous problems¶
Now we will be concerned with the solution of a general problem of the form:
where \(a,\,b,\,c\) are arbitrary constants.
14.2.1. The method of undetermined coefficients¶
This method is a fancy name for educated guesswork - it requires us to pose a trial solution (an ansatz), based on our observation of what \(f(x)\) looks like. We will only be able to apply this method for some specific types of function \(f(x)\), we will be restricted to polynomials, exponentials and trigonometric functions. The method gets its name because we guess the form of the solution without giving the values of the coefficients. We find the values of the undetermined coefficients by substituting our guess into the ODE and equating terms.
Method of undetermined coefficients
Polynomial Form:
Exponential Form:
If we find the that homogeneous solution has a term of the form \(y = A_1\, e^{\alpha x}\). If (and only if) we encounter this problem, you can fix it by multiplying our ansatz by the differentiation variable, here \(x\). The reason this works is similar to the derivation of the linearly independent result in the case of homogeneous problems with a repeated eigenvalue. So the particular integral has the form:
Trigonometric Form:
where we must include both of \(\sin(x)\) and \(\cos(x)\) in our solution.
Hyperbolic Form:
where we must include both of \(\sinh(x)\) and \(\cosh(x)\) in our solution.
Worked example
We will start by considering the following problem:
Since we are trying to make terms like \(e^x\), we will guess a solution of the form \(y_p=Ae^x\). Substituting this guess into the equation gives
We call this function a particular integral. It is not the most general solution to the given problem, as it does not contain any free constants. We will be able to use our particular integral together with the solution to the corresponding homogeneous problem to construct the full result.
Practice questions
Find solutions to the following inhomogeneous problems:
1.
2.
3.
4.
Solutions
1. Here we want to make terms like \(\sin(2x)\) so there will be some of those in our trial guess. However, when we substitute this guess into the ODE we will have some \(\cos(2x)\) terms appearing on the left and so to balance them we will need some of those terms too. Our ansatz is:
Substituting this into the ODE gives:
Equating coefficients of cosine and sine terms gives:
Our particular integral is:
2. Here we want to make terms like \(3x^2+4\) so we will need a trial solution involving \(x^2\) and its derivatives. Taking:
and equating coefficients of \(x^2\), \(x\) and constant terms in the ODE gives:
3. Here we will need an ansatz involving \(e^x\cos(3x)\) and its derivatives. We will take
Substituting in and equating coefficients of \(e^{x}\cos(3x)\) and \(e^x\sin(3x)\) gives:
4. Following the technique that we have used previously, we would think to try the ansatz:
However, if you substitute this into the left hand side of the equation you will find that it simply gives zero. Our desired ansatz here is a solution of the homogeneous problem also - so here we try:
and equating coefficients of \(e^t\) on the left and right sides gives \(A=1\).
14.2.2. Constructing the full solution¶
The general solution to the inhomogeneous problem can be constructed by combining the homogenous solution \(y_h\) and the particular integral \(y_p\):
This works because of the linearity property:
The general solution contains two arbitrary constants so it can be made to satisfy a given set of two conditions. The values of the constants will depend on the particular integral you found. There are an infinite number of different particular integrals possible, but they all will differ only be an amount equal to a linear combination of the basis solutions.
We MUST construct the full solution before employing the given conditions to find constant terms. For example, if you erroneously make \(y_h(0)=y_0\) then we will find mathematically:
which is not what we want.
Worked example
Find the solution to the problem:
The homogenous solution solves:
which given the trial solutions \(y = e^{\lambda x}\) gives:
which means that the solution can be written as:
The particular integral for this problem will have the form:
which means that:
Therefore the general solution is:
Substituting for the initial conditions gives:
The final solution therefore is:
Summary
In order to solve an inhomogeneous problem:
we can apply the following steps:
1. Solve homogeneous problem first, to find \(y_h\), which will include two arbitrary constants:
2. Find a particular integral \(y_p\), by the method of undetermined coefficients:
this should include coefficients that are fixed by the form of the ODE and \(f(x)\).
3. Form a composite solution for \(y(x)\):
in accordance with the principle of superposition of solutions for linear problems.
4. Find constants satisfying any given conditions (this step MUST be done last). e.g. given \(y(0) = \alpha_1,\, y'(0) = \alpha_2\):
14.2.3. A note about the limitations of undeterminded coefficients¶
The method can only be used to solve problems where there is a finite pattern of derivatives for \(f(x)\). So it would not be usable for problems such as:
since the ansatz would need to have an infinite number of terms. To balance the derivative of the logarithm would need a term like \(1/x\), which would need a term like \(1/x^2\) and so forth. There is a method that can be used to solve more general types of inhomogeneous problem - variation of parameters.
14.3. Coupled systems¶
Defintion
Consider again the general form of ODE in (14.2):
We will now introduce a new variable \(z=y^{\prime}\), so that the equation may be rewritten as two:
Notice that we have gone from one second order ODE system to a system of two first order ODEs. The two equations are linked, if \(z\) changes, this will affect \(y\) and vice-versa - we say that they are coupled.
We can write this system in matrix form:
and since the equations are linear, it is also possible to separate out the coefficient matrix:
This sort of system is called an eigenvalue problem and can be solved by techniques from linear algebra.
14.4. Solving separable PDEs¶
Once we extend calculus to many variables, ordinary differential equations (ODEs) can become partial differential equations (PDEs) and then the range of techniques that can be applied to solve these sorts of problems becomes vast - typically there is not clean, closed form, way of solving a given PDE - a series solution or answer valid only in certain regimes is often the best we can do. There is however one class of PDE that can solved using ODE techniques, so called separable PDEs.
Separable PDEs
A separable PDE is of the form:
where \(\lambda\) is some constant and we use the linear differential operators:
We can separate \(u(x,\,t)\) into:
such that the resulting PDE has the form:
where we notice we have switched to one variable derivatives, since:
If we look at (14.5), we notice that if divide each side by \(u = X\,T\):
then notice that whilst the variable \(t\) changing could change the functions on the RHS, it cannot change the value of the LHS, since that does not depends on \(t\). Likewise although variable \(x\) changes the LHS, the RHS would be constant since it does not depend on \(x\), hence we can argue that both sides are equal to some constant, here labelled \(s\):
We can solve this system to find two homogeneous ODE systems:
The size and sign of each of \(a_3,\, b_3,\, \lambda,\, s\) can determine whether there are oscillating or decaying solutions for \(X(x)\) and \(T(t)\).
Worked example
Lets try to solve the wave equation:
where \(c\) is some constant. Using the separable ansatz:
this means that the PDE reduces to:
where \(-k^2\) is used as the separation constant, the choice of which means we pick oscillating solutions, such that the ODEs resulting are:
which each have solutions:
and hence \(u(x,\,t)\) is given by:
Practice question
1. Find solutions to the PDE:
assuming oscillating solutions in \(x,\, t\) and where \(\lambda\) is some constant.
Solution
1. Using the separable ansatz:
this means that the PDE reduces to:
where \(ik\) is used as the separation constant, the choice of which means we pick oscillating solutions, such that the ODEs resulting are:
which each have solutions:
and hence \(u(x,\,t)\) is given by:
Note that we could have picked \(-ik\) as a separation constant, giving:
which is also a valid solution.
14.5. Variation of parameters method¶
Whilst the method of undetermined coefficients works for specific set of source terms \(f(x)\) (i.e. those of certain elementary functions with symmetries in their derivatives), we could ask what is the more general method for solving these sorts of ODEs?
Definition
The variation of parameters method can be used to solve \(n^{th}\) degree linear ODEs of the form:
which can be shown to have the solution:
where \(u_i(x),\, i \in \{1,\, 2,\, \dots,\, n\}\) are the the homogeneous solutions of the ODE, \(W(x)\) is the Wronskian of these homogeneous solutions:
and \(W_i(x)\) is the reduced Wronskian:
where the \(i^{th}\) column of the Wronskian \(W(x)\) has been replaced with:
14.5.1. First order system¶
If we start with a general first order linear ODE:
we can think of this a sourced first order problem, so the homogenous equation is:
which we can solve, for instance with separation of variables:
where \(y_0\) is a constant.
Now we can use this homogeneous solution \(y_h(x)\) to solve for the inhomogeneous case \(y_p(x)\), by multiplying this solution with a (yet unknown) function \(C(x)\):
If we substitute this back into (14.6) we will find:
which by straight integration means that:
and so the final solution is given by the sum of:
which we have seen before with integrating factors for first order systems in (13.3).
14.5.2. Second order system¶
Let’s start with a general form of a second order ODE system::
firstly we aim to solve the homorgeneous equation:
which will admit solutions \(u_1(x),\, u_2(x)\). Hence we will construct a solution to the general equation of the form:
We note that if \(A(x),\, B(x)\) here were just constants, then this would be a linear superposition of the homogeneous solutions - therefore in general these should contain some additive constant with each function. In order to have two equations with two unknowns, we need to enforce a separate condition on \(A(x),\, B(x)\), as we will see the most helpful condition to simplify later algebra turns out to be:
If we try to constuct the form of (14.1) using (14.7), we will find:
where we have used our condition (14.8) to reach the final line. Likewise for second derivatives:
and so using the linear derivative operator:
where we have used the fact that the terms with \(A(x),\, B(x)\) coefficients are actually just the homogeneous equations, hence they equal zero. So this leaves a coupled ODE system:
which if we write as a matrix system \(W(x)\,a = b\):
where \(W(x)\) is the Wronskian of the system.
So to find the solutions \(a = W^{-1}\,b\) we can invert this matrix as:
which means the final solutions can be found from:
Worked example
Lets try and solve the ODE:
subject to the conditions \(y(0) = -\frac{7}{15}, \,y'(0)=\frac{1}{15}\). We note that this could be solved with method of undetermined coefficients, but this means we can double check our results later!
The solutions to the homoegenous ODE can be found from the ansatz \(y = e^{\lambda\,x}\):
which means that in order to construct the full solution:
We need the Wronskian \(W(x)\) of the system, here given by:
and so using (14.9):
wher \(C_1,\, C_2\) are constants. This means the final solution is given by:
Given the initial conditions, we find that:
which we can solve as \(C_1 = 1/2,\, C_2 = -1/2\), which gives final solutions as:
which we can check gives the right answer as:
and so the solutions are correct!