2 The general solution of a differential equation

Consider the equation $y=A{\text{e}}^{2x}$ where $A$ is an arbitrary constant. If we differentiate it we obtain

$\frac{dy}{dx}=2A{\text{e}}^{2x}$

and so, since $y=A{\text{e}}^{2x}$ we obtain

$\frac{dy}{dx}=2y.$

Thus a differential equation satisfied by $y$ is

$\frac{dy}{dx}=2y.$

Note that we have eliminated the arbitrary constant.

Now consider the equation

$y=Acos3x+Bsin3x$

where $A$ and $B$ are arbitrary constants. Differentiating, we obtain

$\frac{dy}{dx}=-3Asin3x+3Bcos3x.$

Differentiating a second time gives

$\frac{d^{2}y}{d{x}^2}=-9Acos3x-9Bsin3x.$

The right-hand side is simply $\left(-9\right)$ times the expression for $y$ . Hence $y$ satisfies the differential equation

$\frac{d^{2}y}{d{x}^2}=-9y.$

Task!

Find a differential equation satisfied by $y=Acosh2x+Bsinh2x$ where $A$ and $B$ are arbitrary constants.

Answer

We have seen that an expression including one arbitrary constant required one differentiation to obtain a differential equation which eliminated the arbitrary constant. Where two constants were present, two differentiations were required. Is the converse true? For example, would a differential equation involving $\frac{dy}{dx}$ as the only derivative have a general solution with one arbitrary constant and would a differential equation which had $\frac{d^{2}y}{d{x}^2}$ as the highest derivative produce a general solution with two arbitrary constants? The answer is, usually, yes.

Task!

Integrate twice the differential equation

$\frac{d^{2}y}{d{x}^2}=\frac{w}{2}\left(\ell x-x^2\right),$

where $w$ and $\ell$ are constants, to find a general solution for $y$ .

Answer

Consider the simple differential equation

$\frac{dy}{dx}=2x.$

On integrating, we obtain the general solution

$y=x^2+C$

where $C$ is an arbitrary constant. As $C$ varies we get different solutions, each of which belongs to the family of solutions. Figure 2 shows some examples.

Figure 2

It can be shown that no two members of this family of graphs ever meet and that through each point in the $x$ - $y$ plane passes one, and only one, of these graphs. Hence if we specify the boundary condition $y=2$ when $x=0$ , written $y\left(0\right)=2$ , then using $y=x^2+c$ :

$2=0+C\text{so that}C=2$

and $y=x^2+2$ is the unique solution.

Task!

Find the unique solution of the differential equation $\frac{dy}{dx}=3x^2$ which satisfies the condition $y\left(1\right)=4$ .

Answer

Example 1

Solve the differential equation $\frac{d^{2}y}{d{x}^2}=6x$ subject to the conditions

1. $y\left(0\right)=2$ and $y\left(1\right)=3$
2. $y\left(0\right)=2$ and $y\left(1\right)=5$
3. $y\left(0\right)=2$ and $\frac{dy}{dx}=1$ at $x=0$ .

Solution

1. Integrating the differential equation once produces  $\frac{dy}{dx}=3x^2+A.$  The general solution is found by integrating a second time to give  $y=x^3+Ax+B$ , where $A$ and $B$ are arbitrary constants.

   Imposing the conditions $y\left(0\right)=2$ and $y\left(1\right)=3$ : at $x=0$ we have $y=2=0+0+B=B$ so that $B=2$ , and at $x=1$ we have $y=3=1+A+B=1+A+2$ . Therefore $A=0$ and the solution is

   $y=x^3+2.$

2. Here the second condition is $y\left(1\right)=5$ so at $x=1$

   $y=5=1+A+2\text{so that}A=2$

   and the solution in this case is

   $y=x^3+2x+2.$

3. Here the second condition is

   $\frac{dy}{dx}=1\text{at}x=0\text{i.e.}{y}^{\prime }\left(0\right)=1$

   then since $\frac{dy}{dx}=3x^2+A$ , putting $x=0$ we get:

   $\frac{dy}{dx}=1=0+A$

   so that $A=1$ and the solution in this case is $y=x^3+x+2$ .