1.3 Linear first-order ODEs

1.3.1 Terminology

Recalling our definition of a linear operator,

L(y)=c_{n}(x)y^{(n)}+c_{n-1}(x)y^{(n-1)}+\ldots+c_{1}(x)y^{\prime}+c_{0}(x)y

(where the $c_{i}(x)$ are specified functions of $x$ ), any linear first-order ODE can be written in the form

equation (1.6) (1.6)

L(y)\equiv c_{1}(x)y^{\prime}+c_{0}(x)y=f(x),

where $f(x)$ is another given function of $x$ . Without loss of generality, we shall assume $c_{1}(x)\neq 0$ .

Note.

The case of $c_{1}(x)=0$ is trivial, i.e. $\displaystyle{c_{0}(x)y=f(x)\Longrightarrow y=\frac{f(x)}{c_{0}(x)}}$ .

Dividing (1.6) by $c_{1}(x)$ ,

equation (1.7) (1.7)

y^{\prime}+p(x)y=q(x),

where $p(x)=c_{0}(x)/c_{1}(x)$ and $q(x)=f(x)/c_{1}(x)$ . This is the most general form for a (non-trivial) first-order linear ODE.

1.3.2 The integrating factor method

The challenge is to solve $y^{\prime}+p(x)y=q(x)$ . We shall assume that $p(x)$ and $q(x)$ are continuous (and hence integrable) over some interval $(a,b)$ , possibly the whole real line $(-\infty,\infty)$ . Let

P(x)=\int p(x)\,\mathrm{d}x.

For any $x\in(a,b)$ , the fundamental theorem of calculus tells us that

P^{\prime}(x)=p(x).

This implies that

\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\left[y\mathrm{e}^{P(x)}\right]

\displaystyle=

\displaystyle y^{\prime}\mathrm{e}^{P(x)}+P^{\prime}(x)y\mathrm{e}^{P(x)}

\displaystyle=

\displaystyle y^{\prime}\mathrm{e}^{P(x)}+p(x)y\mathrm{e}^{P(x)}

\displaystyle=

\displaystyle\left[y^{\prime}+p(x)y\right]\mathrm{e}^{P(x)}.

The quantity $\displaystyle{\mathrm{e}^{P(x)}\equiv\exp\left[P(x)\right]}$ is known as the integrating factor. As we have just shown, if we multiply our original ODE by this integrating factor, i.e.

\left[y^{\prime}+p(x)y\right]\mathrm{e}^{P(x)}=q(x)\mathrm{e}^{P(x)},

this is equivalent to

equation (1.8) (1.8)

\frac{\mathrm{d}}{\mathrm{d}x}\left[y\mathrm{e}^{P(x)}\right]=q(x)\mathrm{e}^{% P(x)}.

Integrating this equation,

y\mathrm{e}^{P(x)}=C+\int\left[q(x)\mathrm{e}^{P(x)}\right]\,\mathrm{d}x,

where $C$ is a constant. Hence, recalling that $\displaystyle{P(x)=\int p(x)\,\mathrm{d}x}$ ,

equation (1.9) (1.9)

y=C\mathrm{e}^{-P(x)}+\mathrm{e}^{-P(x)}\int\left[q(x)\mathrm{e}^{P(x)}\right]% \,\mathrm{d}x.

This is the general solution. The term proportional to $C$ is often referred to as the complementary function, whilst the rest of the right-hand side is known as the particular integral.

Example 1.6.

Find the general solutions of the following first-order ODEs:

i)	$\displaystyle{y^{\prime}+\frac{y}{x}=x}$ ;
ii)	$\displaystyle{5xy^{\prime}+6y=1+x^{2}}$ .

Solution.

i) This equation is already in the form $y^{\prime}+p(x)y=q(x)$ . Here, $p(x)=1/x$ and $q(x)=x$ . Clearly,

P(x)=\int p(x)\,\mathrm{d}x=\int\frac{\mathrm{d}x}{x}=\ln|x|

(note that the integrating factor is simply a computational tool, so we can ignore any constants of integration here). Hence,

\mathrm{e}^{P(x)}=\mathrm{e}^{\ln|x|}=|x|.

Without loss of generality, we can ignore the modulus operator here, setting $\displaystyle{\mathrm{e}^{P(x)}=x}$ (this is because a negative integrating factor would also work in the same way). Multiplying our original equation by this integrating factor, we obtain the following ODE:

x\left(y^{\prime}+\frac{y}{x}\right)=x^{2}\Longrightarrow xy^{\prime}+y=x^{2}.

Referrring to (1.8), we have already shown that this is equivalent to

\frac{\mathrm{d}}{\mathrm{d}x}\left[y\mathrm{e}^{P(x)}\right]=q(x)\mathrm{e}^{% P(x)}\Longrightarrow\frac{\mathrm{d}}{\mathrm{d}x}\left[xy\right]=x^{2}.

Integrating this equation,

\displaystyle xy

\displaystyle=

\displaystyle C+\frac{x^{3}}{3}\quad\mbox{(where $C$ is constant)}

\displaystyle\Longrightarrow y

\displaystyle=

\displaystyle\frac{C}{x}+\frac{x^{2}}{3},

which is the required result.

[Check: $\displaystyle{y^{\prime}+\frac{y}{x}=-\frac{C}{x^{2}}+\frac{2x}{3}+\frac{C}{x^% {2}}+\frac{x}{3}=x}$ , as required.]

ii) This equation ( $\displaystyle{5xy^{\prime}+6y=1+x^{2}}$ ) is not in the correct form to apply the integrating factor directly. Dividing through by $5x$ ,

y^{\prime}+\frac{6y}{5x}=\frac{1}{5x}+\frac{x}{5},

so $p(x)=6/5x$ and $q(x)=(1/5x)+(x/5)$ . Here,

P(x)=\int p(x)\,\mathrm{d}x=\int\frac{\mathrm{6d}x}{5x}=\frac{6}{5}\ln|x|=\ln% \left(|x|^{6/5}\right).

Hence,

\mathrm{e}^{P(x)}=|x|^{6/5}.

Again we can drop the modulus operator, so that $\displaystyle{\mathrm{e}^{P(x)}=x^{6/5}}$ . Applying this integrating factor,

x^{6/5}y^{\prime}+\frac{6yx^{1/5}}{5}=x^{6/5}\left(\frac{1}{5x}+\frac{x}{5}% \right),

which, by (1.8), is equivalent to

\frac{\mathrm{d}}{\mathrm{d}x}\left[x^{6/5}y\right]=\frac{x^{1/5}}{5}+\frac{x^% {11/5}}{5}.

Integrating each side of this equation,

\displaystyle x^{6/5}y

\displaystyle=

\displaystyle C+\frac{x^{6/5}}{6}+\frac{x^{16/5}}{16}\quad\mbox{(where $C$ is % constant)}

\displaystyle\Longrightarrow y

\displaystyle=

\displaystyle Cx^{-6/5}+\frac{1}{6}+\frac{x^{2}}{16},

which is the required solution.

[Check: $\displaystyle{5xy^{\prime}+6y=-6Cx^{-6/5}+\frac{5x^{2}}{8}+6Cx^{-6/5}+1+\frac{% 3x^{2}}{8}=1+x^{2}}$ , as required.]

1.3.3 Existence and uniqueness

Theorem: If the functions $p(x)$ and $q(x)$ are continuous $\forall x\in(a,b)$ , then there exists a unique function $y(x)$ , defined on the open interval $a<x<b$ , satisfying the linear first-order ODE

y^{\prime}+p(x)y=q(x),\quad y\left(x_{0}\right)=y_{0},

for some given $x_{0}\in(a,b)$ and $y_{0}\in\mathbb{R}$ .

Proof: Existence has already been demonstrated by the construction of an explicit solution (1.9). For uniqueness, we assume that there are two solutions $y_{1}(x)$ and $y_{2}(x)$ . Consider the function $Y(x)=y_{1}(x)-y_{2}(x)$ . Clearly

\displaystyle Y^{\prime}+p(x)Y

\displaystyle=

\displaystyle\left(y_{1}-y_{2}\right)^{\prime}+p(x)\left(y_{1}-y_{2}\right)

\displaystyle=

\displaystyle\left[y_{1}^{\prime}+p(x)y_{1}\right]-\left[y_{2}^{\prime}+p(x)y_% {2}\right]

\displaystyle=

\displaystyle q(x)-q(x)

\displaystyle\Longrightarrow Y^{\prime}+p(x)Y

\displaystyle=

\displaystyle 0,

whilst $\displaystyle{Y\left(x_{0}\right)=y_{1}\left(x_{0}\right)-y_{2}\left(x_{0}% \right)=y_{0}-y_{0}=0}$ . Obviously $Y=0$ is a possible solution of this ODE, in which case $y_{1}(x)=y_{2}(x)$ (i.e. the solution is unique). For $Y\neq 0$ , separation of variables leads to

Y^{\prime}=-p(x)Y\iff\int\frac{\mathrm{d}Y}{Y}=-\int p(x)\,\mathrm{d}x\iff\ln|% Y|=C-P(x),

(where, as before. $\displaystyle{P(x)=\int p(x)\,\mathrm{d}x}$ and $C$ is a constant). Rearranging this expression, we see that

Y=A\mathrm{e}^{-P(x)},

where $A$ is a redefined constant. The only way that this can be compatible with $Y(0)=0$ is if $A=0\Longrightarrow Y(x)=0\;\forall x$ . This again implies that $y_{1}(x)=y_{2}(x)$ , i.e. unique solution.

Note.

Linearity is crucial to this result. There are many nonlinear ODEs for which solutions are not unique. For example, consider the following differentiable function,

y(x)=\left\{\begin{array}[]{l}0\hskip 59.98422pt\mbox{if}\hskip 7.22743ptx\leq C% \\ (x-C)^{2}\hskip 21.68121pt\mbox{if}\hskip 7.22743ptx>C,\end{array}\right\}

where $C$ is any positive constant (i.e. $C\geq 0$ ). For any valid choice of $C$ (of which there are infinitely many), this function satisfies the nonlinear ODE

y^{\prime}=2\sqrt{y},

subject to $y(0)=0$ .