10. Second Order ODEs¶

Intended Learning Outcomes

Correctly use the term “homogeneous”
Determine the general solution of any homogeneous second order ODE with constant coefficients
Apply given conditions to determine the constants of integration

10.1. Introduction to the problem¶

This section of the notes will be concerned with the solution of a general problem of the form

(10.1)¶\[ay^{\prime\prime}(x)+by^{\prime}(x)+cy(x)=f(x),\]

where $a,b,c$ are arbitrary constants.

First, we consider the case when $f(x)=0$ so that we obtain the homogeneous problem

(10.2)¶\[ay^{\prime\prime}(x)+by^{\prime}(x)+cy(x)=0\]

We can also conveniently write these problems using a linear operator:

\[\mathcal{L}=a\frac{\mathrm{d}^2}{\mathrm{d}x^2}+b\frac{\mathrm{d}}{\mathrm{d}x}+c\]

With this definition, the problems may be written as

$\mathcal{L}(y)=0$ (homogeneous)
$\mathcal{L}(y)=f(x)$ (inhomogeneous)

10.1.1. An observation¶

If we try a solution of the form $y=e^{\lambda x}$ then the homogeneous problem reduces to $e^{\lambda x}(a\lambda^2+b\lambda +c)=0$ and so by solving the quadratic in $\lambda$ we will have found a solution. We call the quadratic the auxiliary equation. The nature of the solutions depends on the quantity $\Delta= -b^2-4ac$, which we call the discriminant.

Example:
For the problem $2y^{\prime\prime}(x)-y^{\prime}(x)-y(x)=0$, the auxiliary equation is $2\lambda^2-\lambda-1=0$ The quadratic has roots $\lambda=-\frac{1}{2}$ and $\lambda=1$, so we have found two particular solutions of the given ODE, which are $y_1=e^{-\frac{1}{2}x}, \quad y_2=e^{x}$

10.1.2. 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:

(10.3)¶\[\begin{align}\mathcal{L}(k_1y_1+k_2y_2)&=k_1\left[a y_1^{\prime\prime}+by_1^{\prime}+c y_1\right]+k_2\left[a y_2^{\prime\prime}+by_2^{\prime}+c y_2\right]\\&=k_1\mathcal{L}(y_1)+k_2\mathcal{L}(y_2)\\&=0+0 \quad \text{ if $y_1$ and $y_2$ are solutions.}\end{align}\]

Therefore, in the previous example we can combine the two solutions that we found, to obtain a more general solution $y=k_1e^{-\frac{1}{2}x}+k_2 e^{x}$. You are encouraged to check this solution to confirm that it works!

10.1.3. Applying conditions¶

Since the result we obtained has 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.

Example

For the previous problem $2y^{\prime\prime}(x)-y^{\prime}(x)-y(x)=0$

(a) find the particular solution that satisfies $y(0)=2$, $y^{\prime}(0)=1$

(b) find the particular solution that satisfies $y(0)=3$, $y(1)=e^{-1/2}+2e$

Solutions

The general solution is $y=k_1e^{-\frac{1}{2}x}+k_2 e^{x}$, from which we also obtain $y^{\prime}=\frac{1}{4} k_1 e^{-\frac{x}{2}}+k_2 e^x$

(a) Substituting for the given conditions:

$ y(0)=1\Rightarrow k_1+k_2=2$

$ y^{\prime}(0)=1 \Rightarrow -\frac{1}{2}k_1+k_2=1$

Solving these two equations simultaneously gives $k_1=\frac{2}{3}$, $k_2=\frac{4}{3}$.

The solution satisfying the given conditions is therefore given by $y(x)=\frac{2}{3}e^{-\frac{1}{2}x}+\frac{4}{3}e^x$

(b) Substituting for the given conditions:

$ y(0)=3\Rightarrow k_1+k_2=3$

$ y(1)=e^{-1/2}+2e \Rightarrow k_1e^{-1/2}+k_2e=e^{-1/2}+2e$

Solving these two equations simultaneously gives $k_1=1$, $k_2=2$.

The solution satisfying the given conditions is therefore given by $y(x)=e^{-\frac{1}{2}x}+2e^x$

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.

10.1.4. Complex conjugate roots¶

If the discriminant $b^2-4ac$ is negative, the auxiliary equation will have complex conjugate solutions of the form $\lambda=\alpha\pm i\omega$. In that case, we may wish to write the general solution a bit differently. After all, if we have $y=k_1 e^{(\alpha+i\omega)x}+k_2 e^{(\alpha -i\omega)x}$ then the only way to satisfy real initial conditions would be if the coefficients $k_1$ and $k_2$ are complex, which isn’t nice. We make use of our knowledge of complex numbers (Euler’s theorem) to rewrite:

(10.4)¶\[\begin{align}y&=e^{\alpha x}\left[(k_1+k_2)\cos(\omega x)+i(k_1-k_2)\sin(\omega x)\right]\\&=e^{\alpha x}\left[K_1\cos(\omega x)+K_2\sin(\omega x)\right]\end{align}\]

upon relabelling the constants. This is a much nicer form of the solution, because all parts of it are real, and so we expect that the coefficients $K_1$ and $K_2$ 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 $\omega$.

Example

Find the general solution of the problem $y^{\prime\prime}(x)-2y^{\prime}(x)+2y(x)=0$

Solution

The auxiliary equation $\lambda^2-2\lambda+2=0$ has roots $\lambda=1\pm i$. Therefore the general form of the solution is $y=e^{x}\left[K_1\cos(x)+K_2\sin(x)\right]$

10.1.5. Repeated roots:¶

If the discriminant $b^2-4ac$ is zero, the auxiliary equation will only have one distinct solution $e^{\lambda x}$, where $\lambda=-\frac{b}{2a}$. We will look for another solution of the form

\[\begin{equation*} y=f(x)e^{\lambda x}. \end{equation*}\]

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:

\[\begin{equation*} \mathcal{L}(f e^{\lambda x}) = \mathcal{L}(e^{\lambda x})f+e^{\lambda x}\left[(2a\lambda+b )f^{\prime}+a f^{\prime\prime}\right] \end{equation*}\]

The first term disappears since $e^{\lambda x}$ is a solution, and so to satisfy the problem we require $(2a\lambda+b )f^{\prime}+a f^{\prime\prime} =0$. Using the result for the repeated root $\lambda$ reduces this equation to $f^{\prime\prime}=0$, which has solution $f(x)=Ax+B$. We can choose $A=1$ and $B=0$. Thus, we have found another solution $x e^{\lambda x}$ and we can form the general solution from a linear combination :

\[\begin{equation*} y=(k_1x+k_2)e^{\lambda x}. \end{equation*}\]

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.

Example

Example: For the problem $y^{\prime\prime}(x)-4y^{\prime}(x)+4y(x)=0$,

a. find the general solution

b. Verify by substituting into the ODE that your solution works

c. Find the particular solution that satisfies $y(0)=1$, $y^{\prime}(0)=3$

Solution

a. The auxiliary equation is $\lambda^2-4\lambda+4=0$, with repeated root $\lambda=2$.

Hence, the general solution is given by $y=e^{2x}(k_1x+k_2)$, where $k_1$ and $k_2$ are constants.

c. $y(0)=1\Rightarrow k_1=1$,

$y^{\prime}(0)=3 \Rightarrow 2k_1+k_2=3$

The solution satisfying the given conditions is $y-e^{2x}(x+1)$

10.1.6. Summary:¶

The general solution of the homogeneous second order ODE

\[\begin{equation*} ay^{\prime\prime}(x)+by^{\prime}(x)+cy(x)=0 \end{equation*}\]

is given by

(10.5)¶\[\begin{equation} y=\begin{cases}\begin{array}{lll}k_1e^{\lambda_1 x}+k_2 e^{\lambda_2 x}, &\text{where $\lambda_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$} & \text{if $\Delta >0$}\\e^{\alpha x}(k_1\cos(\omega x)+k_2\sin(\omega x))&\text{where $\alpha=-\frac{b}{2a}$, $\omega =\sqrt{\Delta}$}&\text{if $\Delta < 0$}\\(k_1x+k_2)e^{\lambda x}&\text{where $\lambda=-\frac{b}{2a}$}&\text{if $\Delta=0$}\end{array}\end{cases} \end{equation}\]

where $\Delta=b^2-4ac$.

10.1.7. Existence of a particular solution:¶

The general solution

\[\begin{equation*} y(x)=k_1 y_1(x)+k_2 y_2(x) \end{equation*}\]

can be made to satisfy arbitrary initial conditions $y(x_0)=y_0$, $y^{\prime}(x_0)=y^{\prime}_0$ if and only if

(10.6)¶\[y_1^{\prime}(x_0)y_2(x_0)-y_2^{\prime}(x_0)y_1(x_0)\neq 0\]

This condition is met if $y_2$ is not a constant multiple of $y_1$. We say that the solutions are linearly independent. A more general definition of linear independence will be given in the course on mathematical methods 2.

Proof:
Applying the given conditions gives

(10.7)¶\[\begin{align}y_1(x_0)k_1+y_2(x_0)k_2 &=y_0\\y_1^{\prime}(x_0)k_1+y_2^{\prime}(x_0)&=y_0^{\prime}\end{align}\]

This is a set of two equations in two unknowns ($k_0$ and $k_1$). A unique solution exists if and only if the condition (10.6) is met.

10.1.8. Motivation for the trial solution¶

One way to motivate the trial solution $y=e^{\lambda x}$ is to begin by solving the problem for the special case where $c=0$, which reduces to

\[\begin{equation*} az^{\prime}(x)+bz(x)=0,\quad \text{where } z=y^{\prime} \end{equation*}\]

This problem can be solved by separation and the result is of the form $y=k_1e^{-bx/a}+k_2$.

10.2. Inhomogeneous problems¶

Intended Learning Outcomes

For an inhomogenous second order ODE with constant coefficients:

Find a particular integral using the method of undetermined coefficients
Find the general solution by combining the homogeneous solution and a particular integral
Find the particular solution satisfying given conditions.

10.2.1. Introduction to the problem¶

This section of the notes will be concerned with the solution of a general problem of the form

\[\mathcal{L}(y) = ay^{\prime\prime}(x)+by^{\prime}(x)+cy(x)=f(x),\]

where $a,b,c$ are arbitrary constants.

10.2.2. The method of undetermined coefficients¶

This method is a fancy name for educated guesswork. It requires us to pose an ansatz (trial solution), based on our observation of what $f(x)$ looks like. We will only be able to apply this method for some very limited types of function $f(x)$, mainly restricted to polynomials, exponentials and trigonometric functions. It is best illustrated by example.

Example 1:
We will start by considering the following problem:

\[y^{\prime\prime}+4y^{\prime}+5y=e^x\]

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 $10Ae^x=e^x \Rightarrow y_p=\frac{1}{10}e^x$

We call this 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.

But first let’s practice finding some more particular integrals by the method of undetermined coefficients. 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.

Example 2:

a. $y^{\prime\prime}+4y^{\prime}+5y=\sin(2x)$

b. $y^{\prime\prime}+4y^{\prime}+5y=3x^2+4$

c. $y^{\prime\prime}+4y^{\prime}+5y=e^x\cos(3x)$

Solutions

a. 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 $y=A\sin(2x)+B\cos(2x)$.

Substituting into the ODE gives

$(8A+B)\cos(2x)+(A-8B)\sin(2x)=\sin(2x)$

Equating coefficients of cosine and sine terms gives

$8A+B=0, \quad A-8B=1\qquad \Rightarrow A=\frac{1}{65},\quad B=\frac{-8}{65}$

Our particular integral is $y_p=\frac{1}{65}(\sin(2x)-8\cos(2x))$

b. Here we want to “make” terms like $3x^2+4$ so we will need a trial solution involving $x^2$ and its derivatives.

Taking $y_p=Ax^2+bx+c$ and equating coefficients of $x^2$, $x$ and constant terms in the ODE gives

$y_p=\frac{1}{125}(75x^2-120x+166)$

c. Here we will need an ansatz involving $e^x\cos(3x)$ and its derivatives.

We will take $y_p=e^{x}(A\cos(3x)+B\sin(3x))$.

Substituting in and equating coefficients of $e^{x}\cos(3x)$ and $e^x\sin(3x)$ gives

$y_p=\frac{e^x}{325}(\cos(3x)+18\sin(3x))$

Example 3:

Now let’s try a tricky example:

\[2\ddot{x}-\dot{x}-x=3e^{t}\]

Following the technique that we have used previously, we would think to try the ansatz $y_p=Ae^{t}$. 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!

If (and only if) you encounter this problem, you can fix it by multiplying your ansatz by $x$. The reason this works is similar to the derivation of the linearly independent result that was given in the case of homogeneous problems with a repeated eigenvalue.

Here, we try $y_p=Axe^{t}$ and equating coefficients of $e^t$ on the left and right sides gives $A=1$.

10.2.3. A note about the limitations of variation of parameters:¶

The method can only be used to solve problems where there is a finite pattern of derivatives for $f(x)$. So, for example, it would not be usable for problems where $f(x)=\ln(x)$ 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 $1/x^2$, which would need … [etc].

There is a method that can be used to solve more general types of inhomogeneous problem, which is called variation of parameters. It is slightly beyond the scope of this course, but it uses a similar technique to what was done in the derivation of the homogeneous result for repeated real roots - employing known partial solutions to construct the full solution. It assumes a solution of the form $y=u(x)y_1(x)+v(x)y_2(x)$ where $y_1$ and $y_2$ are the homogeneous basis solutions and the functions $u,v$ are to be determined.

10.2.4. Constructing the full solution¶

The general solution to the inhomogeneous problem can be constructed by combining the homogenous solution $y_h$ and a particular integral $y_p$: $y=y_h+y_p$ It works because of the linearity property:

\[\mathcal{L}(y_h+y_p) = \mathcal{L}(y_h)+\mathcal{L}(y_p) = 0+f(x).\]

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.

You 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 you will find mathematical

\[\begin{equation*} y(0)=y_h(0)+y_p(0) = y_0+y_p(0), \end{equation*}\]

which is not what you want.

Example

Find the particular solution to the problem

$y^{\prime\prime}+4y^{\prime}+5y=e^x, \quad y(0)=1, \quad y^{\prime(0)=2}$

Solution

We already found that the particular integral for this problem is $y=\frac{1}{10}e^x$.

The homogenous solution is $y=e^{-2x}(k_1\cos(x)+k_2\sin(x))$

Therefore the general solution is

$y=e^{-2x}(k_1\cos(x)+k_2\sin(x))+\frac{1}{10}e^x$

Substituting for the initial conditions gives

$k_1+\frac{1}{10}=1$ and $k_2-2k_1+\frac{1}{10}=2$, with solution $k_1=\frac{9}{10}$, $k_2=\frac{37}{10}$

The general solution is $y=\frac{1}{10}e^{-2x}(9\cos(x)+37\sin(x))+\frac{1}{10}e^x$

10.2.5. Summary of technique:¶

Solve homogeneous to find $y_h$, which will include two arbitrary constants
Find a particular integral $y_p$, by educated guesswork (method of undetermined coefficients)
Form composite solution $y=y_h+y_p$ in accordance with superposition principle
Find constants satisfying any given conditions (this step MUST be done last).

Mathematics for Natural Sciences 1

10. Second Order ODEs¶

10.1. Introduction to the problem¶

10.1.1. An observation¶

10.1.2. Principle of linear superposition¶

10.1.3. Applying conditions¶

10.1.4. Complex conjugate roots¶

10.1.5. Repeated roots:¶

10.1.6. Summary:¶

10.1.7. Existence of a particular solution:¶

10.1.8. Motivation for the trial solution¶

10.2. Inhomogeneous problems¶

10.2.1. Introduction to the problem¶

10.2.2. The method of undetermined coefficients¶

10.2.3. A note about the limitations of variation of parameters:¶

10.2.4. Constructing the full solution¶

10.2.5. Summary of technique:¶