# Topic 4 Linear Second Order Differential Equations

## 4.1 The general solution of a linear equation

Consider the linear second order differential equation $y''+p(x)y'+q(x)y=f(x).$ Let $$y_p$$ be a particular solution, and $$y_h$$ the general solution of the associated homogeneous equation $y''+p(x)y'+q(x)y=0.$ Then the function $$y_h+y_p$$ is the general solution of the non-homogeneous equation $$y''+p(x)y'+q(x)y=f(x)$$.

The function $$y_h$$ can be written as $$c_1y_1+c_2y_2$$, where $$y_1$$ and $$y_2$$ are linearly independent solutions of the associated homogeneous equation. To find $$y_1$$ and $$y_2$$, we can use dsolve with the option output=basis.

Example 4.1 Find two linearly independent solutions of the equation $y''+ y=0.$

Solution. We first run commands which provide better display.

PDETools[declare](y(x), prime=x): # Turn ON the enhanced DEdisplay feature
infolevel[dsolve] := 3: # Show more detailed information about computation

Now define the equation

ode31:=diff(y(x), x$2)+y(x)=0 # here x$n = x, x, ..., x, n copies of x

To find the independent solutions, run the following command

dsolve(ode31, y(x), output=basis)

The output is $\color{blue}{[\sin(x), \cos(x)]}$ which means that $$\sin x$$ and $$\cos x$$ are linearly independent solutions.

If the homogeneous linear equation has constant coefficients, then one can also use DETools[constcoeffsols] to find a list of independent solutions.

Exercise 4.1 Find two linearly independent solutions of the equation $y'' - y' - 2y=0.$

To find a particular solution of a nonlinear ODE, or a linear non-homogeneous ODE, one can use the command particularsol(ODE, dependent variable).

Example 4.2 Find a particular solution to the equation $y''-3y'+2y=2.$

Solution. First define the equation

ode32:=diff(y(x), x, x)-3*diff(y(x), x)+2*y(x)=2

To find a particular solution, we can use

with(DETools):
particularsol(ode32, y(x));

or

DETools[particularsol](ode32, y(x))

Exercise 4.2 Find a particular solution to the equation $y''-3y'+2xy=2x^2-3.$

## 4.2 Wronskian of Solutions

The Wronskian of two solutions $$y_1$$ and $$y_2$$ of a linear homogeneous second order differential equation $$y''+p(x)y'+q(x)y=0$$ is defined by $W(y_1, y_2)=y_1y_2'-y_1'y_2$. In Maple, one can use the command Wronskian([y_1, y_2], independent variable, determinant=true) or LinearAlgebra[Determinant](Wronskian([y_1, y_2], independent variable) to find the Wronskian.

Example 4.3 The equation $$x^2y''+ 3xy'-3y=0$$ has two solutions $$y_1=x$$ and $$y_2=x^{-3}$$. Find the Wronskian and determine if the two solutions are linearly independent.

Solution. We first load the package VectorCalculus.

with(VectorCalculus):

Now we calculate the Wronskian.

LinearAlgebra[Determinant](Wronskian([x, x^(-3)], x))

The output is $\color{blue}{-\frac{4}{x^{3}}}.$

Because the Wronskian is not identically zero, the solutions are linearly independent.

Exercise 4.3 The equation $$x^2y''+ 4xy'-4y=0$$ has two solutions $$y_1=x$$ and $$y_2=x^{-4}$$. Find the Wronskian and determine if the two solutions are linearly independent.

## 4.3 Linear Equations with Constant Coefficients

When solving linear equations with constant coefficients, we need to solve equations and system of equations. That can be done by the Maple command solve({equations}, {unknowns}). Moreover, one can add constrains by using assuming together with solve.

Example 4.4 Solve the initial value problem $y''-2y'-3y=0,\qquad y(0)=1, y'(0)=-1.$

Solution. Since this equation is a linear second order equation with constant coefficient, the solution is determined by the roots of the characteristic equation $$r^2-2r-3=0$$. Running the following Maple command will give us the roots.

rts:=solve(r^2-2*r-3=0, r);

The roots are $\color{blue}{3, -1}.$

Since the root are real numbers, the general solution of the differential equation is

yh(x):=c1*exp(rts*x)+c2*exp(rts*x);

The initial conditions impose two equations for the constants c1 and c2. We can calculate $$y'$$ using D(y) and solve the constants by the following command

consts:=solve({yh(0)=1, D(yh)(0)=-1}, {c1,c2});

The output is $\color{blue}{\{{c1} = 0, {c2} = 1\}}.$

So the solution to the initial value problem is $y(x)=e^{-x}$ which can be seen by the maple commands

y(x):=subs(consts, yh(x));
y(x);

Here, we take the approach of solving the problem manually. In Maple, we can use dsolve or constcoeffsols to solve differential equations with constant coefficients.

Exercise 4.4 Solve the initial value problem $y''+3y'+2y=0,\qquad y(0)=3, y'(0)=1.$

## 4.4 Undetermined Coefficients

For some equations that can be written into the form $ay''+by'+cy=f(x)$, a particular solution can be found using the method of undetermined coefficients. Calculations can be made easy using Maple.

Example 4.5 Find a particular solution of $y''-5y'+6y=3e^{2x}.$

Solution. Since the right hand side is an exponential function, and 2 is a root of the characteristic polynomial of the complementary equation, we expect a specific solution $$y_p=Axe^{2x}$$. We then plug it into the equation to solve for $$A$$.

Define the differential equation and the solution function.

ode341:=diff(y(x), x, x)-5*diff(y(x), x)+6*y(x)=3*exp(2*x);
yp(x):=A*x*exp(2*x);

Plugging $$y_p$$ into the differential equation and solve for A. Since $$A$$ is a number such that the equation is true for any $$x$$. We use the solve/identity method. The command runs like solve(identity(equation, x), vars).

eqnA:=subs[eval](y(x)=yp(x), ode341);
A=solve(identity(eqnA, x), A);

The output is $\color{blue}{A = -3}.$

Instead of using the solve/identity scheme, one can also build systems of equations for the undetermined coefficients manually.

When $$f(x)$$ involves a polynomial, we want to get a system of undetermined coefficients. One way is to substitute $$x$$ by some general numbers. Another way is to differentiate both sides and then plug in $$x=0$$. Once we get a system of the unknowns, we can solve it by the Maple command solve({eq1, eq2, ...}, {unknown1, unknown2, ...}).

When $$f(x)$$ involves sine or cosine, surely, we can plug in some general values for $$x$$ to obtain a system. We may also substitute $$\sin x$$ by $$0$$ and $$\cos x=0$$ to get two equations. From those two equations, we can build a system.

Example 4.6 Find a particular solution of $y''- 2y'-3y=x\sin x.$

Solution. We expect a specific solution of the same type as the right hand side. Let’s try $$y_p=(ax+b)\cos x+(cx+d)\sin x$$. We then plug it into the equation to solve for $$a$$, $$b$$, $$c$$, and $$d$$.

Define the differential equation and the solution function.

ode342:=diff(y(x), x, x)-2*diff(y(x), x)-3*y(x)=x*sin(x);
yp(x):=(a*x+b)*cos(x)+(c*x+d)*sin(x);

Plugging $$y_p$$ into the differential equation and solve for A.

eqnB:=subs[eval](y(x)=yp(x), ode342);

Now let’s deduce a system of equations for $$a$$, $$b$$, $$c$$ and $$d$$.

# get an equation from coefficients of cosine
CoefCosine := subs([sin(x) = 0, cos(x) = 1], eqnB);

# get an equation from coefficients of sine
CoefSine := subs([sin(x) = 1, cos(x) = 0], eqnB);

# Get four equations for a, b, c, and d
eq1 := subs(x = 0, CoefCosine):
eq2 := subs(x = 0, CoefSine):
eq3 := diff(CoefCosine, x):
eq4 := diff(CoefSine, x): 

Now we solve the equations as a system.

abcd:=solve({eq1, eq2, eq3, eq4},{a, b, c, d});

The output is $\color{blue}{\left\{a = \frac{1}{10}, b = -\frac{7}{50}, c = -\frac{1}{5}, d = -\frac{1}{50}\right\} }.$

Let’s verify that the solution.

Sol:=subs(abcd, yp(x)) # plug a, b, c, d in yp(x)
odetest(Sol, ode342)

If $$f(x)$$ is a polynomial, one can also use the command polysol to find a polynomial solution.

Exercise 4.5 Find a particular solution of $y''- 5y'- 6y=2e^{3x}.$

## 4.5 Reduction of Order

Given a solution $$y_1$$ of a linear second order differential equation $$y'' + p(x)y' + q(x)y=f(x)$$, one can use the trick of variation of parameter to find another solution in the form $$y_2=y_1v$$, here $$v$$ satisfies a second order equation in the form $$y_1v'' + v'(p(x)y_1+2y_1)=0$$. This equation can be solve by reduction of order using the substitution $$u=v'$$. Indeed, we can expect $$y_2=y_1\int u(x)\mathrm{d}x$$.

In Maple, one can use the command reduce_order(ode, y(x)=y_1(x), u(t)) to obtain a lower order equation for $$u$$. This command again belongs to the package DETools.

Example 4.7 The equation $$x^2y''+ 4xy'-4y=0$$ has a solution $$y_1=x$$. Find another linearly independent solution $$y_2$$.

Solution. Let’s load packages first. Since with can only load one package each time, let’s use a for loop to load packages. To make the display looks better, let’s also run the declare command.

for i in [DETools, LinearAlgebra, PDETools] do with(i) end do:
declare(y(x), prime=x):

Define the equation

ode35:=x^2*diff(y(x), x, x)+4*x*diff(y(x),x)-4*y=0;

Find the equation that u satisfies.

RdOrder:=reduce_order(ode35, y(x) = x, u(t));

The output is \color{blue}{ \begin{aligned} y =& \left(t \left({\color{gray}{\int}}u \left(t \right){\color{gray}{d}}t +\_\textit{C1} \right)\right)\boldsymbol{\mathrm{where}}\\ &\left[\left\{t \left(u_{t}\right)+6 u \left(t \right)=0\right\}, \left\{t =x , u \left(t \right)=\frac{x \left({y'}\right)-y}{x^{2}}\right\}, \left\{x =t , y =t \left({\color{gray}{\int}}u \left(t \right){\color{gray}{d}}t +\_\textit{C1} \right)\right\}\right] \end{aligned} }

You can see that the equation for $$u$$ is $$t \left(u_{t}\right)+6 u \left(t \right)=0$$, where $$u_t$$ is $$v'(t)$$.

We can solve for $$u$$ and then find $$v$$. Here, we use the operand extract command op to get the differential equation for $$u$$. The equation is the first element in the first list of the second vector of the right hand side of the output. We now get the equation for $$u(x)$$.

odeu:=subs(t=x, op([2, 1, 1], rhs(RdOrder)));

Now we substitute $$t$$ by $$x$$ and solve for $$u$$ and $$v$$.

ux:=dsolve(odeu, u(x));
v:=int(ux, x);

Therefore, by the following command, we know that $$y_2(x)=\frac{c}{x^4}$$.

y(x):=x*v(x);

Testing $$y_2$$ using the following command shows that it is a solution.

odetest(y(x)=y(x), ode35);
• To get the second solution $$y_2$$ after obtained the differential equation for $$u$$, one can also use the command

buildsol(Reduced_Order_Equation, Particular_Solution_of_u)
• One can also find $$y_2$$ without using DETools (see for example the Maple document on Linear Equations created by Prof. Douglas B. Meade.)

Exercise 4.6 The equation $$x^2y'' + 3xy' - 3y=0$$ has a solution $$y_1=x$$. Find another linearly independent solution $$y_2$$.

## 4.6 Variation of Parameters

Maple has the command varparam(Solutions, f(x), x) to find the general solution of a linear ODE by the method of variation of parameters, where $$f(x)$$ is the right-hand side function. Again this command belongs to DETools.

Example 4.8 Find the generals solution for the equation $x^2y'' + 4xy' - 4y=x$ using the solutions $$y_1=x$$ and $$y_2=\frac{1}{x^4}$$ of the complementary equation $x^2y'' + 4xy' - 4y=0.$

Solution. We can find the general solution using one line of command.

DETools[varparam]([x, 1/x^4], x, x)

The output is $\color{blue}{ \_C_{1} x +\frac{\_C_{2}}{x^{3}}+\frac{x^{3}}{14} }$

Exercise 4.7 Find the generals solution for the equation $x^2y'' + 3xy' - 3y=x$ using the solutions $$y_1=x$$ and $$y_2=\frac{1}{x^3}$$ of the complementary equation $x^2y'' + 3xy' - 3y=0.$