Theory

Introduction

A first-order (explicit) ODE looks like this:

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

for some function $f$.

Many techniques for solving first-order ODEs come from reverse-engineering results of calculus. In this respect, solving ODEs analytically is like taking integrals (finding antiderivatives) analytically. In fact, antidifferentiation is itself one of the techniques: when $f(x,y)$ does not depend on $y$, both sides of the equation above can be integrated to find $y(x)$. More advanced methods like the product rule or $u$-substitution yield more sophisticated techniques for ODEs.

Direct integration

A differential equation of the form:

$$\frac{dy}{dx}=f(x)$$

has a set of solutions given by $y=F(x)+c$ where $c$ is any constant and $F$ is an antiderivative of $f$, meaning $F^{\prime }=f$.

This kind of differential equation is really a part of basic calculus. A few things to emphasize en route to more complicated ODEs:

• There is a family of solutions parametrized by some constant $c$. (This will become a pattern.)
• This family is a complete set: there are no other solutions. (Why not?)
• A particular solution is determined (selected from this family) by a choice of initial condition, which consists of a point $(x_0,y_0)$ that should lie on the solution.
• Consider that the derivative value specifies the direction of travel. So the initial condition is a starting point, and $f(x)$ gives the data of which way to go, changing course along the route.

This technique can be applied in a series for simple ODEs of higher order. For example, the ODE

$$\frac{d^{2}y}{d{x}^2}=f(x)$$

is solved by taking two antiderivatives. Let ${\Phi }^{\prime }=F$ and $F^{\prime }=f$. Then we have a family of solutions that depends on two constants:

$$y(x)=\Phi (x)+c_{1}x+c_2.$$

To specify a particular solution, one provides two points $(x_1,y_1)$ and $(x_2,y_2)$ that should lie on the solution, and you can solve for $c_1$ and $c_2$ by evaluating $\Phi (x_1)$ and $\Phi (x_2)$.

Direct integration (part 2)

Instead of providing multiple points lying on the solution, one can choose a single value $x_1$, and then specify the values of the higher derivatives $y^{(i)}(x_1)=b_i$ using some numbers $b_0,\dots ,b_{n-1}$.

Separable equations - substitution

Consider again the general form of an explicit first-order ODE:

$$\frac{dy}{dx}=f(x,y).$$

This ODE is called separable when $f(x,y)=g(x)h(y)$ for some product of functions of the separate variables, $g(x)$ and $h(y)$. It is solved by integrating separately in $x$ and in $y$, as we consider next.

Consider any function $a(y)$ with antiderivative $A(y)$, so $A^{\prime }=a$. The chain rule says:

$$\frac{d}{dx}A(y)=A^{\prime }(y)\frac{dy}{dx}=a(y)\frac{dy}{dx}.$$

Now let $a(y)=\frac{1}{h(y)}$. The differential equation

$$\frac{dy}{dx}=g(x)h(y)$$

is equivalent to the equation

$$a(y)\frac{dy}{dx}=g(x),$$

so we end up with

$$\frac{d}{dx}A(y)=g(x),$$

which is solved by integrating both sides in $x$ to obtain:

$$A(y)=G(x)+c,$$

where $G^{\prime }=g$ is an antiderivative of $g$ and $c$ is a constant of integration.

Think about this equation $A(y)=G(x)+c$. It equates the antiderivatives of $1/h(y)$ and $g(x)$ (up to a constant). It is an implicit definition of $y(x)$, because it may be virtually impossible to solve for $y$ here as a function of $x$. And, for the same reason as used for the ODE $y^{\prime }(x)=f(x)$ earlier, the family of implicit curves given by this equation (for various $c$) describes all possible solutions.

Many authors describe the solution method a little differently. Starting with

$$\frac{dy}{dx}=g(x)h(y),$$

write it as

$$\frac{dy}{h(y)}=g(x)dx$$

and “integrate” both sides, summing up the differentials:

$$\int \frac{1}{h(y)}dy=\int g(x)dx.$$

This yields

$$A(y)=G(x)+c$$

as before.

This method is easy to remember, and is acceptable to use on an exam or HW, but it is not really standard calculus procedure, since it relies on ambiguous rules about differentials. (Normally in calculus books differentials are considered purely mnemonic devices.)

Linear equations - product rule

First-order linear equations have this general form:

$$y^{\prime }+a(x)y=b(x).$$

These equations are solved using the product rule together with an “integrating factor” $e^{\int a(x)dx}$. It works like this. Apply the product rule to $y\cdot e^{\int a(x)dx}$:

$$\begin{array}{rl}\frac{d}{dx}\left(y\cdot e^{\int a(x)dx}\right)& =y^{\prime }\cdot e^{\int a(x)dx}+y\cdot a(x)e^{\int a(x)dx}\\ & =e^{\int a(x)dx}\left(y^{\prime }+a(x)y\right).\end{array}$$

Now, multiply the original equation on both sides by $e^{\int a(x)dx}$ to find:

$$\frac{d}{dx}\left(y\cdot e^{\int a(x)dx}\right)=e^{\int a(x)dx}\left(y^{\prime }+a(x)y\right)=e^{\int a(x)dx}\cdot b(x).$$

Integrate both sides with respect to $x$:

$$y\cdot e^{\int a(x)dx}=\int e^{\int a(x)dx}\cdot b(x)dx+c,$$

which can be solved for $y$:

$$y=e^{-\int a(x)dx}\int e^{\int a(x)dx}\cdot b(x)dx+c{e}^{-\int a(x)dx}.$$

Exact equations - gradients

A final technique is based on more advanced calculus.

Recall the theorem of vector calculus:

Curl-free fields are gradients

Suppose $\mathbf{F}(x,y)$ is a vector field on ${\mathbb{R}}^2$ with zero curl: $\nabla \times \mathbf{F}=0$. (Assume the region is simply-connected.) Then there is a scalar field $f(x,y)$ whose gradient recovers the vector field: $\mathbf{F}=\nabla f$.

Writing out the components, this theorem says that if

$$\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=0,$$

then there is a function $f$ satisfying

$$F_1=\frac{\partial f}{\partial x}\text{and}{F}_2=\frac{\partial f}{\partial y}.$$

Now, finally, let $\mathcal{C}$ be a level curve given by $f(x,y)=c$ for some $c$. Then $\nabla f$ is perpendicular to $\mathcal{C}$. This latter fact can be interpreted as a differential equation with $\mathcal{C}$ one of its integral curves, as we show next.

Let $y=y(x)$ be considered a function of $x$ whose graph is the curve $\mathcal{C}$. Then $\mathcal{C}$ may be described parametrically by $(x,y(x))$, and the tangent vector to this parametric curve is described parametrically by $(1,y^{\prime }(x))$. The condition $\nabla f\perp \mathcal{C}$ is equivalent to $\nabla f\cdot (1,y^{\prime })=0$, which when written out means:

$$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}{y}^{\prime }=0.$$

Writing $M(x,y)=\frac{\partial f}{\partial x}$ and $N(x,y)=\frac{\partial f}{\partial y}$ and $y^{\prime }=\frac{dy}{dx}$, and multiplying through by $dx$, we obtain a form that is easier to remember:

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

Here is a summary of our reasoning:

Exact equations

Suppose the differential equation written either in form (A):

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

or in form (B):

$$M(x,y)+N(x,y)\frac{dy}{dx}=0$$

satisfies $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. Then it has a family of solutions given by the level curves $f(x,y)=c$, where $f$ is a function satisfying $\frac{\partial f}{\partial x}=M$ and $\frac{\partial f}{\partial y}=N$.

Algebraic summary

A more algebraic formulation of the same theory goes like this:

If $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, then it is possible to find $f(x,y)$ such that $\frac{\partial f}{\partial x}=M$ and $\frac{\partial f}{\partial y}=N$. In this case the equation

$$M(x,y)+N(x,y)y^{\prime }=0$$

can be written

$$\frac{d}{dx}(f(x,y(x)))=0$$

using the chain rule, and this has as solutions the family of integral curves defined implicitly with constants $c\in \mathbb{R}$:

$$f(x,y(x))=c.$$