Theory

Introduction

So far we have seen how to solve differential equations by direct integration, by separation of variables, by integrating factors: conversion to a product (linear equations), and by gradients (exact equations).

Many differential equations do not immediately appear to have the standard form to which these methods are applicable, but in fact these methods can be applied after performing a transformation. The possibility of some of these transformations can be checked systematically, but for others we rely on an art – not a mechanical science – of finding an appropriate substitution. This art is based on experience and honed by practice.

Homogeneous equations - conversion to separable

Homogeneous functions
A function is homogeneous when the input variable appears everywhere in the same degree. More precisely and technically, $f(x)$ is homogeneous when $f(tx)=t^{\alpha }f(x)$ for some $\alpha \in \mathbb{R}$. Similarly $g(x,y)$ is homogeneous when $g(tx,ty)=t^{\alpha }g(x,y)$ for some $\alpha \in \mathbb{R}$. (Likewise for more variables.) In this situation $\alpha$ is called the degree of homogeneity.

In other words, a function is homogeneous when the action of scaling the inputs simultaneously can be factored out to an action of scaling the output.

For example, these functions are homogeneous:

$$x^2+xy,\sin (x/y),\sqrt{x^2+y^2}$$

whereas these are not:

$$x^2+x,\sin (xy),x\sqrt{1-y^2}.$$

Homogeneous differential equations
Now, suppose we have a differential equation in the form:

$$\frac{dy}{dx}=\frac{M(x,y)}{N(x,y)},$$

where $M$ and $N$ are homogeneous functions of the same degree. Then the change of variables $z=y/x$ (by substituting $y=zx$) will convert the quotient on the right to a function of $z$ alone, and the equation as a whole to a separable equation, which we can solve for $z$ and at the end the solution can be converted to an expression in $y$ by substituting back $z=y/x$.

Homogeneous substitution

Suppose $M(x,y)$ and $N(x,y)$ are homogeneous of the same degree $\alpha$. Then $M(x,y)/N(x,y)$ depends only on the ratio $z=y/x$:

$$\frac{M(x,y)}{N(x,y)}=\frac{M(x,zx)}{N(x,zx)}=\frac{x^{\alpha }M(1,z)}{x^{\alpha }N(1,z)}=\frac{M(1,z)}{N(1,z)}=:F(z).$$

Separability then occurs because:

$$\frac{d}{dx}y=\frac{d}{dx}(zx)=\frac{dz}{dx}x+z,$$

so the equation $\frac{dy}{dx}=F(z)$ is equivalent to

$$\frac{dz}{F(z)-z}=\frac{dx}{x},$$

and the general solution is the family of curves given by:

$$\int \frac{dz}{F(z)-z}=\ln |x|+c.$$

Alternate format
The differential equation

$$M(x,y)dx+N(x,y)dy=0$$

is homogeneous if $M(x,y)$ and $N(x,y)$ are both homogeneous of the same degree. It can be transformed with the same substitution $z=y/x$ and the result is separable.

Homogeneous equations - conversion to separable (part 2)

Linear quotients: almost homogeneous
Suppose we are given a differential equation

$$\frac{dy}{dx}=\frac{a_{1}x+b_{1}y+c_1}{a_{2}x+b_{2}y+c_2}.$$

This equation is not homogeneous, but it is easily converted to a homogeneous equation by a change of variables. Find $h,k$ that solve the system:

$$\begin{array}{rl}{a}_{1}h+b_{1}k& =-c_1\\ a_{2}h+b_{2}k& =-c_2.\end{array}$$

Then by plugging in $x=u+h$ and $y=v+k$, the quotient on the right side is transformed to a quotient of homogeneous functions in $u$ and $v$:

$$\frac{dy}{dx}=\frac{a_{1}u+b_{1}v}{a_{2}u+b_{2}v}$$

The left side transforms as desired:

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=\frac{dv}{du}\cdot 1=\frac{dv}{du}.$$

Integrating factors - conversion to exact

Sometimes an equation

$$M(x,y)dx+N(x,y)dy=0$$

is not exact, but there is a function $\mu (x,y)$ which when multiplied through the equation gives an exact equation:

$$\mu Mdx+\mu Ndy=0.$$

Integrating factors - conversion to exact (part 2)

How can we find such a function $\mu (x,y)$? There is no recipe that can find all possibilities for $\mu (x,y)$. However, if there is a $\mu$ that works and is a function of either $x$ or $y$ alone, then there is a recipe. Here it is:

Integrating factors: single variable

• Check $x$: if the function $g(x)=\frac{1}{N}(M_y-N_x)$ depends on $x$ alone, we can use $\mu (x)=e^{\int g(x)dx}$.
• Check $y$: if the function $h(y)=-\frac{1}{M}(M_y-N_x)$ depends on $y$ alone, we can use $\mu (y)=e^{\int h(y)dy}$.

Reduction of order

Some second-order ODEs can be reduced to first-order ODEs by substituting $y^{\prime }=p$. This works when the equation $F(x,y,y^{\prime },y^{\prime \prime })$ is missing either $x$ or $y$.

If the equation is missing $y$, so the ODE is actually given by $F(x,y^{\prime },y^{\prime \prime })=0$, then we substitute $y^{\prime }=p$ and $y^{\prime \prime }=p^{\prime }$, and try to solve the equation $F(x,p,p^{\prime })=0$ as any other first-order equation. Afterwards we can solve for $y$ by integrating $p$ (provided we have an explicit solution for $p(x)$).

If the equation is missing $x$, so the ODE is actually given by $F(y,y^{\prime },y^{\prime \prime })=0$, then we substitute $y^{\prime }=p$ as before, but now we let $p=p(y)$ depend on $y$, and so $y^{\prime \prime }=p^{\prime }(y)p$ by the chain rule:

$$y^{\prime \prime }=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p^{\prime }(y)p.$$

We therefore try to solve the equation $F(y,p,p^{\prime }p)$ as any other first-order equation.

Ad hoc transformations

Bernoulli equations
The Bernoulli equations look like linear equations but with an extra factor of $y^n$:

$$\frac{dy}{dx}+a(x)y=b(x)y^n.$$

If we substitute $v=y^{1-n}$, a Bernoulli equation becomes a linear equation.

Ad hoc transformations (part 2)

Linear substitutions
If in the general form of an explicit first-order equation $y^{\prime }=f(x,y)$ the items $x$ and $y$ only occur together in a linear combination, so it really takes the form $y^{\prime }=f(ax+by)$, then we can change variables using $z=ax+by$ and then separate variables. First convert $y^{\prime }$:

$$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{z(x)-ax}{b}\right)=\frac{z^{\prime }(x)-a}{b},$$

then the equation $y^{\prime }=f(ax+by)=f(z)$ becomes $z^{\prime }=a+bf(z)$, which is separable:

$$\frac{dz}{a+bf(z)}=dx.$$

Swapping roles
It is sometimes effective to transform an equation to a known form by swapping the roles of $x$ and $y$, meaning that $x$ is to be treated as the dependent variable and $y$ as the independent. To do this we need to replace $y^{\prime }$ with $1/x^{\prime }$. This can be explained and justified using the theory for derivatives of inverse functions.

Inverse functions derivatives

If $x=f(y)$, then $1=f^{\prime }(y)\frac{dy}{dx}$ so $\frac{dy}{dx}=1/f^{\prime }(y)$. Now treating $f(y)=x(y)$, we have $f^{\prime }(y)=\frac{dx}{dy}$. Therefore $\frac{dy}{dx}=1/\frac{dx}{dy}$.