2 Finding the complementary function

To find the complementary function we must make use of the following property.

If $y_1\left(x\right)$ and $y_2\left(x\right)$ are any two (linearly independent) solutions of a linear, homogeneous second order differential equation then the general solution $y_{\text{cf}}\left(x\right),$ is

$y_{\text{cf}}\left(x\right)=A{y}_1\left(x\right)+B{y}_2\left(x\right)$

where $A,B$ are constants.

We see that the second order linear ordinary differential equation has two arbitrary constants in its general solution . The functions $y_1\left(x\right)$ and $y_2\left(x\right)$ are linearly independent if one is not a multiple of the other.

Example 5

Verify that $y_1={\text{e}}^{4x}$ and $y_2={\text{e}}^{2x}$ both satisfy the constant coefficient linear homogeneous equation:

$\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+8y=0$

Write down the general solution of this equation.

Solution

When $y_1={\text{e}}^{4x}$ , differentiation yields:

$\frac{d{y}_1}{dx}=4{\text{e}}^{4x}\text{ and}\frac{d^2{y}_1}{d{x}^2}=16{\text{e}}^{4x}$

Substitution into the left-hand side of the ODE gives $16{\text{e}}^{4x}-6\left(4{\text{e}}^{4x}\right)+8{\text{e}}^{4x}$ , which equals 0, so that $y_1={\text{e}}^{4x}$ is indeed a solution.

Similarly if $y_2={\text{e}}^{2x},$ then

$\frac{d{y}_2}{dx}=2{\text{e}}^{2x}\text{and}\frac{d^2{y}_2}{d{x}^2}=4{\text{e}}^{2x}.$

Substitution into the left-hand side of the ODE gives $4{\text{e}}^{2x}-6\left(2{\text{e}}^{2x}\right)+8{\text{e}}^{2x}$ , which equals 0, so that $y_2={\text{e}}^{2x}$ is also a solution of equation the ODE. Now ${\text{e}}^{2x}$ and ${\text{e}}^{4x}$ are linearly independent functions, so, from the property stated above we have:

$y_{\text{cf}}\left(x\right)=A{\text{e}}^{4x}+B{\text{e}}^{2x}$ is the general solution of the ODE.

Example 6

Find values of $k$ so that $y={\text{e}}^{kx}$ is a solution of:

$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-6y=0$

Hence state the general solution.

Solution

As suggested we try a solution of the form $y={\text{e}}^{kx}$ . Differentiating we find

$\frac{dy}{dx}=k{\text{e}}^{kx}\text{and}\frac{d^{2}y}{d{x}^2}=k^2{\text{e}}^{kx}.$

Substitution into the given equation yields:

$k^2{\text{e}}^{kx}-k{\text{e}}^{kx}-6{\text{e}}^{kx}=0\text{ that is}\left(k^2-k-6\right){\text{e}}^{kx}=0$

The only way this equation can be satisfied for all values of $x$ is if

$k^2-k-6=0$

that is, $\left(k-3\right)\left(k+2\right)=0$ so that $k=3$ or $k=-2$ . That is to say, if $y={\text{e}}^{kx}$ is to be a solution of the differential equation, $k$ must be either 3 or $-2$ . We therefore have found two solutions:

$y_1\left(x\right)={\text{e}}^{3x}\text{and}{y}_2\left(x\right)={\text{e}}^{-2x}$

These are linearly independent and therefore the general solution is

$y_{\text{cf}}\left(x\right)=A{\text{e}}^{3x}+B{\text{e}}^{-2x}$

The equation $k^2-k-6=0$ for determining $k$ is called the auxiliary equation .

Task!

By substituting $y={\text{e}}^{kx}$ , find values of $k$ so that $y$ is a solution of

$\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0$

Hence, write down two solutions, and the general solution of this equation.

First find the auxiliary equation:

Answer

Answer

Example 7

Find the auxiliary equation of the differential equation:

$a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0$

Solution

We try a solution of the form $y={\text{e}}^{kx}$ so that

$\frac{dy}{dx}=k{\text{e}}^{kx}\text{and}\frac{d^{2}y}{d{x}^2}=k^2{\text{e}}^{kx}.$

Substitution into the given differential equation yields:

$a{k}^2{\text{e}}^{kx}+bk{\text{e}}^{kx}+c{\text{e}}^{kx}=0\text{that is}\left(a{k}^2+bk+c\right){\text{e}}^{kx}=0$

Since this equation is to be satisfied for all values of $x$ , then

$a{k}^2+bk+c=0$

is the required auxiliary equation.

Task!

Write down, but do not solve, the auxiliary equations of the following:

1. $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$ ,
2. $2\frac{d^{2}y}{d{x}^2}+7\frac{dy}{dx}-3y=0$
3. $4\frac{d^{2}y}{d{x}^2}+7y=0$ ,
4. $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=0$

Answer

Solving the auxiliary equation gives the values of $k$ which we need to find the complementary function. Clearly the nature of the roots will depend upon the values of $a,b$ and $c$ .

Case 1 If $b^2>4ac$ the roots will be real and distinct. The two values of $k$ thus obtained, $k_1$ and $k_2$ , will allow us to write down two independent solutions: $y_1\left(x\right)={\text{e}}^{k_{1}x}\text{and}{y}_2\left(x\right)={\text{e}}^{k_{2}x},$  and so the general solution of the differential equation will be:

$y\left(x\right)=A{\text{e}}^{k_{1}x}+B{\text{e}}^{k_{2}x}$

Case 2 On the other hand, if $b^2=4ac$ the two roots of the auxiliary equation will be equal and this method will therefore only yield one independent solution. In this case, special treatment is required.

Case 3 If $b^2<4ac$ the two roots of the auxiliary equation will be complex, that is, $k_1$ and $k_2$ will be complex numbers. The procedure for dealing with such cases will become apparent in the following examples.

Example 8

Find the general solution of: $\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}-10y=0$

Solution

By letting $y={\text{e}}^{kx}$ , so that $\frac{dy}{dx}=k{\text{e}}^{kx}\text{and}\frac{d^{2}y}{d{x}^2}=k^2{\text{e}}^{kx}$

the auxiliary equation is found to be: $k^2+3k-10=0\text{and so}\left(k-2\right)\left(k+5\right)=0$

so that $k=2$ and $k=-5$ . Thus there exist two solutions: $y_1={\text{e}}^{2x}\text{and}{y}_2={\text{e}}^{-5x}.$

We can write the general solution as: $y=A{\text{e}}^{2x}+B{\text{e}}^{-5x}$

Example 9

Find the general solution of: $\frac{d^{2}y}{d{x}^2}+4y=0$

Solution

As before, let $y={\text{e}}^{kx}$ so that $\frac{dy}{dx}=k{\text{e}}^{kx}\text{and}\frac{d^{2}y}{d{x}^2}=k^2{\text{e}}^{kx}.$

The auxiliary equation is easily found to be:  $k^2+4=0$  that is, $k^2=-4$ so that $k=\pm 2\text{i}$ , that is, we have complex roots. The two independent solutions of the equation are thus

$y_1\left(x\right)={\text{e}}^{2\text{i}x}{y}_2\left(x\right)={\text{e}}^{-2\text{i}x}$

so that the general solution can be written in the form $y\left(x\right)=A{\text{e}}^{2\text{i}x}+B{\text{e}}^{-2\text{i}x}$ .

However, in cases such as this, it is usual to rewrite the solution in the following way.

Recall that Euler’s relations give: ${\text{e}}^{2\text{i}x}=cos2x+\text{i}sin2x\text{and}{\text{e}}^{-2\text{i}x}=cos2x-\text{i}sin2x$

so that $y\left(x\right)=A\left(cos2x+\text{i}sin2x\right)+B\left(cos2x-\text{i}sin2x\right)$ .

If we now relabel the constants such that $A+B=C$ and $A\text{i}-B\text{i}=D$ we can write the general solution in the form:

$y\left(x\right)=Ccos2x+Dsin2x$

Note: In Example 8 we have expressed the solution as $y=\dots$ whereas in Example 9 we have expressed it as $y\left(x\right)=\dots$ . Either will do.

Example 10

Given $a{y}^{″}+b{y}^{\prime }+cy=0,$ write down the auxiliary equation. If the roots of the auxiliary equation are complex (one root will always be the complex conjugate of the other) and are denoted by $k_1=\alpha +\beta \text{i}$ and $k_2=\alpha -\beta \text{i}$ show that the general solution is:

$y\left(x\right)={\text{e}}^{\alpha x}\left(Acos\beta x+Bsin\beta x\right)$

Solution

Substitution of $y={\text{e}}^{kx}$ into the differential equation yields $\left(a{k}^2+bk+c\right){\text{e}}^{kx}=0$ and so the auxiliary equation is:

$a{k}^2+bk+c=0$

If $k_1=\alpha +\beta \text{i},k_2=\alpha -\beta \text{i}$ then the general solution is

$y=C{\text{e}}^{\left(\alpha +\beta \text{i}\right)x}+D{\text{e}}^{\left(\alpha -\beta \text{i}\right)x}$

where $C$ and $D$ are arbitrary constants.

Using the laws of indices this is rewritten as:

$y=C{\text{e}}^{\alpha x}{\text{e}}^{\beta \text{i}x}+D{\text{e}}^{\alpha x}{\text{e}}^{-\beta \text{i}x}={\text{e}}^{\alpha x}\left(C{\text{e}}^{\beta \text{i}x}+D{\text{e}}^{-\beta \text{i}x}\right)$

Then, using Euler’s relations, we obtain:

Writing $A=C+D$ and $B=C\text{i}-D\text{i},$ we find the required solution:

$y={\text{e}}^{\alpha x}\left(Acos\beta x+Bsin\beta x\right)$

Task!

Find the general solution of $y^{″}+2y^{\prime }+4y=0.$

Write down the auxiliary equation:

Answer

Answer

Answer

Example 11

The auxiliary equation of $a{y}^{″}+b{y}^{\prime }+cy=0$ is $a{k}^2+bk+c=0.$ Suppose this equation has equal roots $k=k_1$ and $k=k_1$ . Verify that $y=x{\text{e}}^{k_{1}x}$ is a solution of the differential equation.

Solution

We have: $y=x{\text{e}}^{k_{1}x}{y}^{\prime }={\text{e}}^{k_{1}x}\left(1+k_{1}x\right)y^{″}={\text{e}}^{k_{1}x}\left(k_1^{2}x+2k_1\right)$

Substitution into the left-hand side of the differential equation yields:

${\text{e}}^{k_{1}x}\left\{a\left(k_1^{2}x+2k_1\right)+b\left(1+k_{1}x\right)+cx\right\}={\text{e}}^{k_{1}x}\left\{\left(a{k}_1^2+b{k}_1+c\right)x+2a{k}_1+b\right\}$

But $a{k}_1^2+b{k}_1+c=0$ since $k_1$ satisfies the auxiliary equation. Also,

$k_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

but since the roots are equal, then $b^2-4ac=0$ hence $k_1=-b∕2a$ . So $2a{k}_1+b=0$ . Hence ${\text{e}}^{k_{1}x}\left\{\left(a{k}_1^2+b{k}_1+c\right)x+2a{k}_1+b\right\}={\text{e}}^{k_{1}x}\left\{\left(0\right)x+0\right\}=0$ . We conclude that $y=x{\text{e}}^{k_{1}x}$ is a solution of $a{y}^{″}+b{y}^{\prime }+cy=0$ when the roots of the auxiliary equation are equal. This illustrates Key Point 8.

Example 12

Obtain the general solution of the equation: $\frac{d^{2}y}{d{x}^2}+8\frac{dy}{dx}+16y=0$ .

Solution

As before, a trial solution of the form $y={\text{e}}^{kx}$ yields an auxiliary equation $k^2+8k+16=0$ . This equation factorizes so that $\left(k+4\right)\left(k+4\right)=0$ and we obtain equal roots, that is, $k=-4$ (twice). If we proceed as before, writing $y_1\left(x\right)={\text{e}}^{-4x}$ $y_2\left(x\right)={\text{e}}^{-4x},$ it is clear that the two solutions are not independent. We need to find a second independent solution. Using the result summarised in Key Point 8, we conclude that the second independent solution is $y_2=x{\text{e}}^{-4x}$ . The general solution is then:

$y\left(x\right)=\left(A+Bx\right){\text{e}}^{-4x}$

Exercises

1. Obtain the general solutions, that is, the complementary functions, of the following equations:
   1. $\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+2y=0$
   2. $\frac{d^{2}y}{d{x}^2}+7\frac{dy}{dx}+6y=0$
   3. $\frac{d^{2}x}{d{t}^2}+5\frac{dx}{dt}+6x=0$
   4. $\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}+y=0$  
   5. $\frac{d^{2}y}{d{x}^2}-4\frac{dy}{dx}+4y=0$ $$
   6. $\frac{d^{2}y}{d{t}^2}+\frac{dy}{dt}+8y=0$
   7. $\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+y=0$   
   8. $\frac{d^{2}y}{d{t}^2}+\frac{dy}{dt}+5y=0$  
   9. $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-2y=0$
   10. $\frac{d^{2}y}{d{x}^2}+9y=0$
   11. $\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}=0$   
   12. $\frac{d^{2}x}{d{t}^2}-16x=0$
2. Find the auxiliary equation for the differential equation $L\frac{d^{2}i}{d{t}^2}+R\frac{di}{dt}+\frac{1}{C}i=0$

   Hence write down the complementary function.

3. Find the complementary function of the equation $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$

Answer