Nonlinear second order ODEs

So far we have only looked at second order LODEs. The techniques we have used to solve the ODEs rely on the fact that the principle of superposition may be used to find a general solution. When we depart from linearity, finding general solutions to differential equations becomes a far more difficult task. In general, the second-order, nonlinear ODE (NLODE), may be expressed as,

We can identify some obvious differences between a LODE and NLODE:

1. A NLODE does not possess the property that a linear combination of two (or more, if higher order) linearly independent solutions is also a solution;
2. It is much rarer that a solution to a NLODE may be expressed in exact form. It is therefore necessary to know how to use alternative methods such as:

(a) Qualitative methods (like the method in Section 12.6);

(b) Numerical methods.

3. Generally, if we can solve a NLODE to obtain a solution, the resulting solution is not a general solution; in other words, NLODEs may possess singular solutions. The latter is a solution that cannot be described by the general integral obtained through solving the ODE.

Special cases

Very little is known about second order nonlinear ODEs but, in a few cases, a second order NLODE can be solved as a first order ODE through the use of a substitution. In particular, we look at two cases:

(i) $y^{\prime \prime}=f\left(x, y^{\prime}\right)$, where $y$ is missing;

(ii) $y^{\prime \prime}=f\left(y, y^{\prime}\right)$, where $x$ is missing.

Here, $y=y(x)$ where $x$ is the independent variable and $y$ is the dependent variable. Case (ii) represents second order autonomous ODEs where $f$ does not depend explicitly on the independent variable.

Case 1: $y$ is missing

Such ODEs take the form,

where the substitution $v=y^{\prime}$ clearly transforms Eq. (13.149) to a first order ODE, namely,

Equation (13.150) can be solved to find a solution describing $v(x)$. Then, $y(x)$ is a solution to,

Consider the following ODE,

We define $v=y^{\prime}$ which means $v^{\prime}=y^{\prime \prime}$. In (13.152),

Equation (13.153) is separable and can be solved to give,

Using a $u$-substitution, $x=c_{1} \tan u$, we obtain the following integral describing the solution to the NLODE, $y(x)$,

Simplifying (13.155) and integrating gives,

Case 2: $x$ is missing

In the case where the independent variable is missing, i.e.

we can still solve the NLODE using the same substitution as before but the substitution is used in a different way. We let $y^{\prime}=v$ and differentiate,

where in (13.158), we have made use of the chain rule. The NLODE (13.157) becomes,

Now, we can think of (13.159) as an ODE for $v(y)$, i.e. $y$ is the independent variable and $v$ is the dependent variable. If we can solve for $v(y)$ then $y(x)$ is a solution to,

Consider the following NLODE,

Let $y^{\prime}=v$; from (13.158) we have $y^{\prime \prime}=v \frac{d v}{d y}$. Then, Eq. (13.161) is transformed to,

Separating variables and integrating gives us:

But, $v=y^{\prime}$ so upon integrating once more, the solution to the NLODE (13.161) is:

Table 13.1: 'Simple' forms of $f(x)$ and trial solution for $y_{p}$