Given an ODE in terms of $\frac{dy}{dx}$, where $y$ is called the dependent variable and $x$ is the independent variable, our aim is to solve the ODE and determine the function, or functions, that satisfy the equation.

Consider the simple ODE

$$\frac{dy}{dx} = x^2 - 1.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[1]$$

The solution we desire will be of the form

$$y = f(x).$$

In this case the task is straightforward as we seek the function, $y$, whose derivative is,

$$x^2 - 1.$$

Hence,

$$y = \int (x^2 - 1)\,dx \;\text{and so} $$

$y = \frac{1}{3}x^3 - x + C$.$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[2]$

Note that an arbitrary constant $( C )$ has been generated. This means that expression $[2]$ does not represent just a single solution of ODE $[1]$, but infinitely many ( one for each possible value of $C$ ). Expression [2] is called the general solution (GS) of $[1]$ since it covers all possible solutions of the given ODE.

The order of an ODE is given by the highest derivative appearing in it. For example,

${\Large\frac{dy}{dx}} = x^2 - 1$ 1st order ODE

$a{\Large\frac{d^2x}{dt^2}} + b{\Large\frac{dx}{dt}} + c \; x = 0$ 2nd order ODE

$y{\Large\frac{d^2y}{dx^2}} = \bigg({\Large\frac{dy}{dx}}\bigg)^3$ 2nd order ODE.

The order of an ODE determines the number of arbitrary constants appearing in its general solution:

1st order - 1 arbitrary constant

2nd order - 2 arbitrary constants.

We shall not be considering ODEs of order higher that 2.

Once the general solution of an ODE has been found, a particular solution may be determined by applying boundary or initial conditions. For the simple first order ODE $[1]$ above we might be asked to solve

$$\frac{dy}{dx} = x^2 - 1, \;\text{subject to}\; y(0) = 1. $$

That is, from the infinity of solutions of the ODE we are looking for the one that will give $y = 1$ when $x = 0$. The general solution was found earlier to be

$$y = \frac{1}{3}x^3 - x + C.$$

To determine the particular solution that satisfies the given condition, we determine a value for $C\;$ by substituting $y = 1$ and $x = 0$ into the general solution and solving the resulting equation, i.e.

$$1 = 0 - 0 + C$$

so that

$$C = 1,$$

giving

$y = \frac{1}{3}x^3 - x + 1$

which satisfies both the ODE and the condition.

In general, we require one condition per arbitrary constant to determine a particular solution. Hence, the number of conditions required equals the order of the ODE.

In the simple case above we were able to determine the general solution of ODE [1] by a single, direct integration. Other ODEs may require different solution methods. We shall now consider some special types of ODEs and how to solve them.

An ODE is separable if it can be written in the form

$$f(y)\frac{dy}{dx} = g(x).$$

Integrating both sides, with respect to $x$, we obtain

$$\int f(y)\, \frac{dy}{dx}dx = \int g(x)\,dx$$

or

$$\int f(y)\,dy = \int g(x)\,dx. $$

Provided we can perform the two integrations we can obtain a solution to the ODE.

Examples

In the following examples, Examples (1) – (5) illustrate the basic processes involved when solving separable ODEs. Example (6) illustrates how an initial condition can be applied as soon as the constant of integration appears. Example (7) shows that some solutions cannot be manipulated into the explicit form $y = f(x)$, and so we have to be content with an implicit form of solution.

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Now watch the following:

📹 Differential Equations - Variables Separable 1

📹 Differential Equations - Variables Separable 2

📹 Differential Equations - Variables Separable 3

An ODE is linear if it is linear in the dependent variable and its derivatives. No multiplications involving the dependent variable and its derivatives are allowed. In a linear ODE all coefficients must therefore be constants or functions of the independent variable. An equation that is not linear is nonlinear. Some examples of linear and nonlinear ODEs are:

$x^2 {\Large\frac{dy}{dx}} + y^3 = \cos(x)$ 1st order nonlinear   ( $y^3$ is the nonlinear term )

$x {\Large\frac{dy}{dx}} - y = x^2e^{3x}$ 1st order linear

${\Large\frac{d^2y}{dx^2}} + y{\Large\frac{dy}{dx}} + y = 0$ 2nd order nonlinear   ( $y{\Large\frac{dy}{dx}}$ is the nonlinear term )

$3 {\Large\frac{d^2y}{dx^2}} + 10 {\Large\frac{dy}{dx}} - 8y = \sin(2x)$ 2nd order linear

$e^{3x}{\Large\frac{dy}{dx}} + \sin(y) = 5x^2$ 1st order nonlinear   ( $\sin(y)$ is the nonlinear term ).

Before we look at the next method for solving first order ODEs it will be helpful to recall some useful results:

Laws of Logarithms

$N\, \ln \, (M) = \ln \,(M^N)$

$\ln\, (M\,N) = \ln\,(M) + \ln\, (N)$

$\ln \bigg({\Large\frac{M}{N}}\bigg) = \ln\,(M) - \ln\,(N)$

$e^{\ln(A)} = A$

Integration by Parts

$$\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\, dx.$$

Consider an ODE that can be written in the standard form

$$\frac{dy}{dx} + p(x)y = q(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[3]$$

where $p(x)$ and $q(x)$ are non-zero and not equal. This type of equation is known as a first order linear ODE and can be solved using the integrating factor method. The solution procedure involves the following steps:

1. Write the ODE in standard form.

2. Identify $p(x)$ and determine the integrating factor (IF), $r(x) = e^{\int p(x)\,dx}$.

3. Multiply the standard form ODE [3] by $r(x)$:

$r(x){\Large\frac{dy}{dx}} + r(x)\,p(x)y = r(x)\,q(x)$.

4. Rewrite as, ${\Large\frac{d}{dx}}[r(x)y] = r(x)q(x)$.

5. Integrate both sides, with respect to $x$, to obtain the general solution:

$r(x)y = {\huge\int} r(x)q(x)\,dx \Rightarrow y = {\Large\frac{1}{r(x)}} {\huge\int} r(x)\,q(x)\,dx$.

Note that we can actually omit Steps 3 and 4 and, after calculating the integrating factor at Step 2, write the general solution ( Step 5 ) as, $y = {\Large\frac{1}{r(x)}} {\huge\int}\,r(x)\,q(x)\,dx$.

Solving ODEs of the form [3] using integrating factors means performing two integrations, one to determine the IF and one to eliminate the derivative. It might appear that we end up with two constants of integration when we only want one. However, it is easily demonstrated that any constant arising from calculating $r(x)$ eventually cancels out. In practice, we simply set it to zero as soon as it arises leaving just one arbitrary constant from the final integration. As before, if an initial condition is given then the constant of integration can be evaluated once it appears.

Example 8

Example 9

Now watch the following:

📹 Differential Equations : First Order Linear 1

📹 Differential Equations : First Order Linear 2

In this section we look at special types of second order ODEs. The following definitions will prove useful for the understanding of parts of this section.

Definition (1): Given two functions $y_1(x)$ and $y_2(x)$ a linear combination of these functions takes the form

$y(x) = a_1\,y_1(x) + a_2\,y_2(x)$

where $a_1$ and $a_2$ are constants.

Definition (2): Two functions, $y_1(x)$ and $y_2(x)$ are said to be linearly independent if one function is not a constant multiple of the other, i.e.

$y_1(x) \neq k\, y_2(x)$.

Definition (3): If two functions, $y_1(x)$ and $y_2(x)$ are both solutions of a linear homogeneous ODE then

$y(x) = A\,y_1(x) + B\, y_2(x)$,

where $A$ and $B$ are constants, is also a solution of the ODE.

Linear ODEs with constant coefficients come in two forms:

$a{\Large\frac{d^2y}{dx^2}} + b{\Large\frac{dy}{dx}} + c\;y = 0$ (Homogeneous) $[4]$

$a{\Large\frac{d^2y}{dx^2}} + b{\Large\frac{dy}{dx}} + c\;y = f(x)$ (Non-homogeneous) $[5]$

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

Equations of the above form are extremely useful for modelling certain types of physical systems. Consider a simple suspension system linking a mass to some fixture via a spring and dashpot:

In this system:

$m =$ mass

$k =$ spring stiffness

$c =$ damping coefficient

$x =$ displacement of mass

${\Large\frac{dx}{dt}} =$ velocity of mass

${\Large\frac{d^2x}{dt^2}} =$ acceleration of mass

Force on mass due to spring $= -k\;x$

Force on mass due to dashpot $= -c{\Large\frac{dx}{dt}}$.

Newton's 2nd law of motion states that Mass $\times$ Acceleration $=$ Force, so

$m{\Large\frac{d^2x}{dt^2}} = -k\;x - c{\Large\frac{dx}{dt}}$

or

$m{\Large\frac{d^2x}{dt^2}} + c{\Large\frac{dx}{dt}} + k\;x = 0$. ( c.f. Equation $[4]$ )

If there are other forces acting on the mass (such as a forced vibration) then we obtain

$m{\Large\frac{d^2x}{dt^2}} + c{\Large\frac{dx}{dt}} + k\;x = f(t)$ ( c.f. Equation $[5]$ )

where $f(t)$ is called the forcing term.

Similar ODEs can be derived that model the behaviour of current or voltage in electronic circuits. These type of equations are applicable to many areas where oscillations or vibrations are present.

6.1. Homogeneous Equations

Consider the second order, linear, homogeneous ODE with constant coefficients

$a{\Large\frac{d^2y}{dx^2}} + b {\Large\frac{dy}{dx}} + c\;y = 0$. $[6]$

To obtain a solution $( y )$ of this type of ODE we first note that in order to satisfy $[6]$ some linear combination of the function $y$ and its derivatives must equal zero. One class of function for which this property holds is the exponential function as differentiation corresponds to multiplication of the previous derivative by some constant.

We shall therefore consider a solution $y$ of the form

$y = e^{\lambda \,x}$

where $\lambda$ is a constant. If we differentiate this function twice we obtain

${\Large\frac{dy}{dx}} = \lambda\;e^{\lambda \, x}$

and

${\Large\frac{d^2y}{dx^2}} = \lambda ^2\,e^{\lambda \,x}$.

Note that $y$ and its derivatives are constant multiples of each other and have a common factor of $e^{\lambda \,x}$. If we substitute these expressions into ODE [6] we obtain

$(a \, \lambda^2 + b\,\lambda + c)\,e^{\lambda \, x} = 0$

or

$a\,\lambda^2 + b \, \lambda + c = 0$, $[7]$

since $e^{\lambda \, x} \neq 0$. If we choose $\lambda$ to satisfy this quadratic equation then, automatically, the function $y = e^{\lambda \, x}$ will be a particular solution of the ODE.

Equation [7] is called the auxiliary equation ( AE ), or characteristic equation, of the ODE. Solving it allows us to construct particular solutions and, hence, the general solution of the ODE. Since we have to solve a quadratic equation we have to deal separately with the different types of solution that can occur. Recall that the quadratic formula gives the solution of the AE as

$\lambda = {\Large\frac{-b\, \pm\, \sqrt{b^2\,-\, 4\,a\,c}}{2\,a}}$.

and so we must consider different cases depending on the value of the discriminant, $b^2 - 4\,a\,c$.

Case (i) $\;\;b^2 - 4\,a\,c > 0$

If $b^2 - 4\,a\,c > 0$ we obtain two real and distinct roots of the AE, i.e.

$\lambda = \lambda_1$  and  $\lambda = \lambda_2$

Hence,

$y_1 = e^{\lambda_1 \, x}$

will be one solution of the ODE and

$y_2 = e^{\lambda_2x}$

will be another. These solutions are linearly independent and by Definition (3) above any linear combination of them will also be a solution of the ODE, i.e.

$y = A\,e^{\lambda_1\,x} + B\;e^{\lambda_2\,x}$ $[8]$

will be a solution. In fact, expression [8] covers all possible solutions of the ODE (when $b^2 - 4\,a\,c > 0$ ) and is therefore the general solution ( GS ) of the ODE. Note the two arbitrary constants. We have constructed the general solution without the need to integrate.

Case (ii) $\;\;b^2 - 4\,a\,c < 0$

In this case we obtain two complex conjugate roots of the auxiliary equation, i.e.

$\lambda_1 = \alpha + j \, \beta$   and   $\lambda_2 = \alpha - j \, \beta$.

Just as in case (i) the general solution can be written down as

$y = \tilde{A}\,e^{\lambda_1 \, x} + \tilde{B}\,e^{\lambda_2 \, x}$

or

$y = \tilde{A}\,e^{(\alpha \,+\, j\,\beta)x} + \tilde{B}\,e^{(\alpha \,-\, j \, \beta)x}$

or

$y = \tilde{A}\,e^{\alpha \, x}\,e^{j \, \beta\,x} + \tilde{B}\,e^{\alpha\,x}\,e^{-\,j\,\beta\,x}$

or

$y = e^{\alpha\,x}\big[\tilde{A}\,e^{j\,\beta\,x} + \tilde{B}\,e^{-\,j\,\beta\,x}\big]$.

Using Euler’s formula we can write

$e^{j\,\beta\,x} = \cos(\beta\,x) + j\,\sin(\beta\,x)$

and

$e^{-\,j\,\beta\,x} = \cos(\beta\,x) - j\,\sin(\beta\,x)$.

Hence, the general solution of the ODE can be rewritten as

$y = e^{\alpha\,x} \big[\tilde{A}\, [\cos(\beta\,x) + j\, \sin(\beta\,x)] + \tilde{B}[\cos(\beta\,x) - j\,\sin(\beta\,x)]\big]$

or

$y = e^{\alpha\,x}\big[A\, \cos(\beta\,x) + B\, \sin(\beta\,x)]$$[9]$

where $A = (\tilde{A} + \tilde{B})$ and $B = j(\tilde{A} - \tilde{B})$.

Equation $[9]$ provides us with a template solution when the auxiliary equation yields complex conjugate roots.

Case (iii) $\;\;b^2 - 4\,a\,c = 0$

In this case the roots of the auxiliary equation are real and equal, i.e.

$\lambda = \lambda_0$.

Both roots give the same solution

$y = e^{\lambda_0\,x}$

However, we need to find two linearly independent solutions to the ODE in order to determine the general solution.

In this special case it can be shown that

$y = e^{\lambda_0\,x}$$[10]$

and

$y = x\,e^{\lambda_0\,x}$$[11]$

are both linearly independent solutions of the ODE and so taking a linear combination of [10] and [11] gives us the general solution

$y = A\,e^{\lambda_0\,x} + B\,x\,e^{\lambda_0\,x}$

or

$y = (A + B\,x)\,e^{\lambda_0\,x}$.

Summary

To solve the ODE

$$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + c\,y = 0 :$$

(I). form the auxiliary equation $a\,\lambda^2 + b\,\lambda + c = 0$;

(II). solve the auxiliary equation;

(III). write down the general solution of the ODE, depending on the nature of the solution of the auxiliary equation:

Two real distinct roots: $\;y = A\,e^{\lambda_1\,x} + B\,e^{\lambda_2\,x}$

Two complex roots: $\;\;\;\;\;\,y = e^{\alpha\,x}\,[A\,\cos(\beta\,x) + B\,\sin(\beta\,x)]$

Two real equal roots: $\;\;\;\,y = (A + B\,x)\,e^{\lambda_0\,x}$.

Now watch the following:

📹 2nd Order Linear Differential Equations - Auxillary Equation : Introduction

Example 10

Example 11

Example 12

6.1.1. Homogeneous ODEs with Initial Conditions

Since the general solution of a 2nd order ODE has two arbitrary constants we require two initial conditions to obtain a particular solution. These conditions may be given in the form

$y(a) = c_0$ ${\Large\frac{dy}{dx}}(a) = c_1$

where the second condition is applied to the derivative of the general solution. In general, we will obtain two simultaneous equations which are solved for $A$ and $B$ to completely solve the ODE. This type of problem, when all the constraints are specified at the same value of the independent variable $(x)$, is known as an initial value problem (IVP).

Example 13

Example 14

6.2. Non-Homogeneous Equations

In this section we look at how to solve second order, linear, non-homogeneous ODEs with constant coefficients, i.e. equations of the form

$a{\Large\frac{d^2y}{dx^2}} + b{\Large\frac{dy}{dx}} + c\,y = f(x)$ $[12]$

where $a$ , $b$ and $c$ are constants

The general solution of a non-homogeneous ODE of the form $[12]$ is found as follows:

STEP 1 Determine the general solution of the corresponding homogeneous ODE (i.e. initially set the RHS, $f(x)$, to zero). In this context, we call the solution the complementary function (CF) and denote it $y_c(x)$.

STEP II Determine a particular integral (PI), which is any solution of the full non-homogeneous ODE, by the method of undetermined coefficients (see below) and denote it $y_p(x)$.

STEP III The general solution of the full, non-homogeneous equation is then given by

$y(x) = y_c(x) + y_p(x)$.

6.2.1. The Method of Undetermined Coefficients

The method enables the derivation of a particular integral, $y_p(x)$, based on the form of the non-homogeneous term, $f(x)$, in the ODE. While the method only works for a limited set of $f(x)$ terms ( polynomials, exponentials, sines and cosines and combinations of these functions ) it is relatively straightforward to use and only involves algebra and basic differentiation. The method is best illustrated by an example. Consider the ODE:

${\Large\frac{d^2y}{dx^2}} - {\Large\frac{dy}{dx}} - 2\,y = x^2 + 4$. $[13]$

Step I: Calculate the complementary function:

AE: $\lambda^2 - \lambda - 2 = 0$

$(\lambda - 2)(\lambda + 1) = 0$

$\lambda = 2 \;\;\;\;\; \lambda = -1$

Two real distinct roots, so the complementary function is,

$y_c(x) = A\,e^{2\,x} + B\,e^{-x}$.

Step II: The aim is to determine a particular integral of the full, non-homogeneous equation. Its format depends on the type of function, $f(x)$, on the RHS of the ODE. Looking at the RHS above we see (in this case) a quadratic. It is therefore reasonable to assume (make an ‘educated guess’) that a particular integral will be of a similar form, namely a quadratic. We therefore write down the most general form that a quadratic can take,

$y_p = a\,x^2 + b\,x + c$, $[14]$

where $a$ , $b$ and $c$ are the, as yet, undetermined coefficients. Our task is now to calculate the values of $a$ , $b$ and $c$ that make [14] a solution of ODE [13]. To do this substitute the general form $[14]$, and its first and second derivatives, into the LHS of ODE [13] and force this new LHS to equal the given RHS.

We have

${\Large\frac{dy_p}{dx}} = 2\,a\,x + b$ $[15]$

${\Large\frac{d^2y_p}{dx^2}} = 2\,a$. $[16]$

Substituting $[14]$, $[15]$ and $[16]$ into the LHS of $[13]$ we get

$[2\,a] - [2\,a\,x + b] - 2[a\,x^2 + b\,x + c] = x^2 + 4$.

Collecting together the powers of $x$ on the LHS gives

$(2\,a -b - 2\,c) - (2\,a + 2\,b)\,x - 2\,a\,x^2 = x^2 + 4.$ $[17]$

If $[14]$ is to be a solution of $[13]$ then we must determine $a$ , $b$ and $c$ so that the LHS of $[17]$ equals the RHS. Equating the coefficients of the powers of $x$ on both sides yields three equations in $a$ , $b$ and $c$ :

coefficients of $x^2$ :$\;\;\;\;\;\;\;\;\;\;-2\,a = 1$

coefficients of $x$ : $-(2\,a + 2\,b) = 0$

constant term : $\;\;\;2\,a - b - 2\,c = 4$ .

Solving these equations gives

$a = -{\Large\frac{1}{2}}, \;\;\;\;\;\; b = {\Large\frac{1}{2}}, \;\;\;\;\;\; c = -{\Large\frac{11}{4}}$.

A particular integral is therefore

$y_p(x) = -{\Large\frac{1}{2}}x^2 + {\Large\frac{1}{2}}x - {\Large\frac{11}{4}}$.

Step III: We can now write down the general solution of ODE $[13]$ by adding together the complementary function, $y_c(x)$, and the particular integral, $y_p(x)$ i.e.

$y(x) = y_c(x) + y_p(x)$

so that

$y(x) = A\,e^{2\,x} + B\,e^{-x} - {\Large\frac{1}{2}}x^2 + {\Large\frac{1}{2}}x - {\Large\frac{11}{4}}$.

For other types of RHS functions, $f(x)$ , we look for particular integrals which are similar in form and the table on the next page gives the trial function (‘educated guess’) for $y_p(x)$ corresponding to various RHS functions. The method can run into difficulty however if the non-homogeneous term, $f(x)$, forms part of the complementary function. This situation is addressed in Example 16.

See Examples (15) – (16) after the table of $y_p$ starting forms.

Table of Trial Functions for Particular Integrals of Non-homogeneous ODEs

RHS function $f(x)$	Trial function for $y_p(x)$
Any constant term ( e.g. $4$ , $-3$ )	$a$
Any linear term ( e.g. $2 x$ , $5 x - 8$ )	$a\,x + b$
Any quadratic term ( e.g. $3 x^2 + 2 x - 6$ )	$ax^2 + bx + c$
RHS contains an exponential of the form $e^{k\,x}$ (where $k$ is not a root of the AE)	$a\,e^{k\,x}$
RHS contains an exponential of the form $e^{k\,x}$ (where $k$ is one of two distinct roots of the AE)	$a\,x\,e^{k\,x}$
RHS contains an exponential of the form $e^{k\,x}$ (where $k$ is a repeated root of the AE)	$a\,x^2\,e^{k\,x}$
RHS contains $\cos(k\,x)$ or $\sin(k\,x)$ or both	$a\,\cos(k\,x) + b\,\sin(k\,x)$ or $x \,[a \,\cos(k\,x) + b\,\sin(k\,x)\,]\;\;$*

* This only occurs with ODEs of the form

${\Large\frac{d^2y}{dx^2}} + k^2y =\;$ RHS containing $\cos(k\,x)$ or $\sin(k\,x)$ or both.

Example 15a

Example 15b

Example 15c

Now watch the following:

📹 2nd Order Linear Differential Equations : P.I. = Constant

📹 2nd Order Linear Differential Equations : P.I. = Linear type

📹 2nd Order Linear Differential Equations : P.I. = Quadratic type

📹 2nd Order Linear Differential Equations : P.I. = Exponential type

📹 2nd Order Linear Differential Equations : P.I. = Trig type

If the non-homogeneous term $f(x)$ appears in the complementary function the method described above will fail. The following examples demonstrate how to overcome this problem.

Example 16a

Example 16b

Example 16c

6.2.2. Non-Homogeneous ODE with Initial Conditions

Note that if we are given an ODE with initial conditions we can calculate the values of $A$ and $B$ and completely solve the ODE. This type of problem, when all the constraints are specified at the same value of the independent variable $(x)$, is known as an initial value problem (IVP). In this case we must derive the full general solution before applying the initial conditions to find $A$ and $B$.

Example 17