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2.3 Linear Equations: Class Notes

2.3a

Definition

A first-order differential equation of the form ${\displaystyle a_1(x)\frac{dy}{dx}+a_0(x)y=g(x)}$ is a linear equation in $y$ (the dependent variable). This is put into standard form by dividing both sides by $a_1(x)$ to obtain

$${\displaystyle \frac{dy}{dx}+P(x)y=f(x)}$$

The function $g(x)$ is called the input function or forcing function. A term such as $yy\prime$ would make an equation nonlinear.

Example

Consider the DE ${\displaystyle 3\frac{dy}{dx}+12y=4}$. In standard form, this is ${\displaystyle \frac{dy}{dx}+4y=\frac{4}{3}}$.

If we multiply both sides by $e^{4x}$, we get

$${\displaystyle e^{4x}\frac{dy}{dx}+4e^{4x}y=\frac{4}{3}{e}^{4x}}$$

Aside: differentiate $e^{4x}y$ with respect to $x$. First, a second aside to recall how this works.

In the same way, by the product rule,

$${\displaystyle \frac{d}{dx}\left[e^{4x}y\right]=e^{4x}\frac{dy}{dx}+4e^{4x}y}$$

This is exactly the left side of the equation above, so the equation can be written as ${\displaystyle \frac{d}{dx}\left[e^{4x}y\right]=\frac{4}{3}{e}^{4x}}$. Integrate both sides with respect to $x$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \int \frac{d}{dx}\left[e^{4x}y\right]dx}& {\displaystyle =\int \frac{4}{3}{e}^{4x}dx}\hfill \\ \hfill {\displaystyle e^{4x}y}& {\displaystyle =\frac{1}{3}{e}^{4x}+C}\hfill \end{array}}$$

$e^{4x}$ is called an integrating factor.

Where did $e^{4x}$ come from?

For the equation in standard form

$${\displaystyle \frac{dy}{dx}+P(x)y=f(x),(*)}$$

suppose we have some function $\mu (x)$ ($\mu$ is the Greek letter mu) such that when we multiply both sides of $(*)$ by $\mu (x)$, we get

$${\displaystyle \mu \frac{dy}{dx}+\mu Py=\mu f(x)}$$

If it happens that ${\displaystyle \frac{d\mu }{dx}=\mu P}$, then the left side becomes

$${\displaystyle \mu \frac{dy}{dx}+\frac{d\mu }{dx}y=\mu f(x)}$$

and by the product rule, ${\displaystyle \mu y\prime +\mu \prime y=\frac{d}{dx}(\mu y)}$. Then we have

$${\displaystyle \frac{d}{dx}(\mu y)=\mu f(x).}$$

To find $\mu$, solve the separable equation

$${\displaystyle \frac{d\mu }{dx}=\mu P\Rightarrow \int \frac{d\mu }{\mu }=\int P(x)dx\Rightarrow \ln |\mu |=\int P(x)dx}$$

Tip: A constant of integration is not necessary when finding or using an integrating factor: ${\displaystyle e^{\int Pdx+C_1}=e^{C_1}{e}^{\int Pdx}}$, and any constant multiple of an integrating factor works just as well.

2.3b

Example

Find the general solution of the given differential equation. Give the largest interval $I$ over which the general solution is defined. Determine whether there are any transient terms in the general solution.

$$y\prime +2xy=x^3$$

Solution

This is a linear DE, so use an integrating factor. $P(x)=2x$, so

$${\displaystyle \mu =e^{\int P(x)dx}=e^{\int 2xdx}=e^{x^2}}$$

Multiply both sides by $e^{x^2}$. The left side has the form $\mu y\prime +\mu \prime y$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle e^{x^2}y\prime +2x{e}^{x^2}y}& {\displaystyle =x^3{e}^{x^2}}\hfill \\ \hfill {\displaystyle \frac{d}{dx}\left(e^{x^2}y\right)}& {\displaystyle =x^3{e}^{x^2}}\hfill \\ \hfill {\displaystyle e^{x^2}y}& {\displaystyle =\int x^3{e}^{x^2}dx}\hfill \end{array}}$$

Integrate by parts with $u=x^2$, $dv=x{e}^{x^2}dx$, so $du=2xdx$, ${\displaystyle v=\frac{1}{2}{e}^{x^2}}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle e^{x^2}y}& {\displaystyle =\frac{1}{2}{x}^2{e}^{x^2}-\int x{e}^{x^2}dx}\hfill \\ \hfill {\displaystyle e^{x^2}y}& {\displaystyle =\frac{1}{2}{x}^2{e}^{x^2}-\frac{1}{2}{e}^{x^2}+C}\hfill \end{array}}$$

${\displaystyle \frac{1}{2}{x}^2-\frac{1}{2}}$ is the steady-state solution. The term $C{e}^{-x^2}$ goes to $0$ as $x\to \infty$, so it is a transient term.

Interval: $(-\infty ,\infty )$. Transient term: $C{e}^{-x^2}$.

Example

$${\displaystyle (1+x)\frac{dy}{dx}-xy=x+x^2}$$

Solution

Standard form: divide by $1+x$, using $x+x^2=x(1+x)$:

$${\displaystyle \frac{dy}{dx}-\frac{x}{1+x}y=x}$$

So ${\displaystyle P(x)=-\frac{x}{1+x}}$. Write the numerator as $x+1-1$ to split the fraction:

$${\displaystyle -\int \frac{x+1-1}{1+x}dx=-\int (1-\frac{1}{x+1})dx=\ln |x+1|-x}$$

$$\mu =e^{\ln |x+1|-x}=e^{\ln |x+1|}{e}^{-x}=(x+1)e^{-x}$$

(Here we take $x>-1$, so $|x+1|=x+1$.) Multiply by $\mu$ and integrate:

$${\displaystyle \int \frac{d}{dx}[(x+1)e^{-x}y]dx=\int (x^2+x)e^{-x}dx}$$

Integrate by parts with $u=x^2+x$, $dv=e^{-x}dx$, so $du=(2x+1)dx$, $v=-e^{-x}$:

$${\displaystyle (x+1)e^{-x}y=-(x^2+x)e^{-x}+\int (2x+1)e^{-x}dx}$$

Integrate by parts again with $u=2x+1$, $dv=e^{-x}dx$, so $du=2dx$, $v=-e^{-x}$:

$$(x+1)e^{-x}y=-(x^2+x)e^{-x}-(2x+1)e^{-x}-2e^{-x}+C$$

Combine $-(2x+1)-2=-(2x+3)$ and divide by $(x+1)e^{-x}$:

No transient terms. The solution is defined on $(-\infty ,-1)\cup (-1,\infty )$; the interval of solution is $(-1,\infty )$.

2.3c

Example

$$xy\prime +(1+x)y=e^{-x}\sin 2x$$

Solution

Standard form (divide by $x$):

$${\displaystyle y\prime +(\frac{1}{x}+1)y=\frac{\sin 2x}{x{e}^x}}$$

$${\displaystyle \mu =e^{\int (\frac{1}{x}+1)dx}=e^{\ln x+x}=x{e}^x}$$

Multiplying by $x{e}^x$ turns the right side into ${\displaystyle x{e}^x·\frac{\sin 2x}{x{e}^x}=\sin 2x}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}\left[x{e}^{x}y\right]}& {\displaystyle =\sin 2x}\hfill \\ \hfill {\displaystyle x{e}^{x}y}& {\displaystyle =-\frac{1}{2}\cos 2x+C}\hfill \end{array}}$$

Interval of solution: $(0,\infty )$. It's all transient: both terms go to $0$ as $x\to \infty$.

Example

$${\displaystyle \frac{dP}{dt}+2tP=P+4t-2}$$

Solution

Here the independent variable is $t$ and the dependent variable is $P$. Move the $P$ term to the left to get standard form:

$${\displaystyle \frac{dP}{dt}+(2t-1)P=4t-2}$$

$${\displaystyle \mu =e^{\int (2t-1)dt}=e^{t^2-t}}$$

Multiply both sides by $e^{t^2-t}$. The left side has the form $\mu P\prime +\mu \prime P$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle e^{t^2-t}P\prime +(2t-1)e^{t^2-t}P}& {\displaystyle =(4t-2)e^{t^2-t}}\hfill \\ \hfill {\displaystyle \frac{d}{dt}\left[e^{t^2-t}P\right]}& {\displaystyle =(4t-2)e^{t^2-t}=2(2t-1)e^{t^2-t}}\hfill \end{array}}$$

Integrate with the substitution $u=t^2-t$, so $du=(2t-1)dt$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \int \frac{d}{dt}\left[e^{t^2-t}P\right]dt}& {\displaystyle =\int 2(2t-1)e^{t^2-t}dt}\hfill \\ \hfill {\displaystyle e^{t^2-t}P}& {\displaystyle =2e^{t^2-t}+C}\hfill \end{array}}$$

Transient term: $C{e}^{t-t^2}$. Interval: $(-\infty ,\infty )$.

Example

$${\displaystyle (x^2-1)\frac{dy}{dx}+2y=(x+1{)}^2}$$

Solution

The equation is linear if its standard form is ${\displaystyle \frac{dy}{dx}+P(x)y=f(x)}$. Divide by $x^2-1=(x+1)(x-1)$:

$${\displaystyle \frac{dy}{dx}+\frac{2}{x^2-1}y=\frac{x+1}{x-1}}$$

So ${\displaystyle P(x)=\frac{2}{x^2-1}}$. To find ${\displaystyle \int P(x)dx}$, use partial fractions:

$${\displaystyle \frac{2}{(x+1)(x-1)}=\frac{A}{x+1}+\frac{B}{x-1}\Rightarrow 2=A(x-1)+B(x+1)}$$

Setting $x=-1$ gives $A=-1$. Setting $x=1$ gives $B=1$.

$${\displaystyle \mu =e^{\int (\frac{1}{x-1}-\frac{1}{x+1})dx}=e^{\ln |x-1|-\ln |x+1|}=e^{\ln \left|\frac{x-1}{x+1}\right|}=\frac{x-1}{x+1}}$$

(On the interval $(1,\infty )$, ${\displaystyle \frac{x-1}{x+1}}$ is positive, so the absolute value can be dropped.) Multiply the standard form by $\mu$:

$${\displaystyle \frac{x-1}{x+1}\frac{dy}{dx}+\underset{\mu \prime }{\underbrace{\frac{x-1}{x+1}·\frac{2}{x^2-1}}}y=\frac{x-1}{x+1}·\frac{x+1}{x-1}}$$

The left side is ${\displaystyle \mu y\prime +\mu \prime y=\frac{d}{dx}[\mu y]}$ and the right side simplifies to $1$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \int \frac{d}{dx}\left[\frac{x-1}{x+1}y\right]dx}& {\displaystyle =\int 1dx}\hfill \\ \hfill {\displaystyle \frac{x-1}{x+1}y}& {\displaystyle =x+C}\hfill \end{array}}$$

No transient terms. Interval: $(1,\infty )$.

2.3d

Example

Solve the given initial-value problem. Use a graphing utility to graph the continuous function $y(x)$.

$${\displaystyle (1+x^2)\frac{dy}{dx}+2xy=f(x),y(0)=0,\text{where}}$$

$$f(x)=\{\begin{array}{cc}{\displaystyle x,}\hfill & {\displaystyle 0\le x<1}\hfill \\ {\displaystyle -x,}\hfill & {\displaystyle x\ge 1}\hfill \end{array}$$

Solution

(A graphing utility is a program such as Desmos or GeoGebra.) Solve the DE separately on each piece of $f$. Standard form on each piece:

$${\displaystyle 0\le x<1:\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{x}{1+x^2}}$$

$${\displaystyle x\ge 1:\frac{dy}{dx}+\frac{2x}{1+x^2}y=-\frac{x}{1+x^2}}$$

Both pieces have the same integrating factor:

$${\displaystyle \mu =e^{\int \frac{2x}{1+x^2}dx}=e^{\ln (1+x^2)}=1+x^2}$$

For $0\le x<1$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}\left[(1+x^2)y\right]}& {\displaystyle =x}\hfill \\ \hfill {\displaystyle (1+x^2)y}& {\displaystyle =\frac{1}{2}{x}^2+C_1}\hfill \\ \hfill {\displaystyle y}& {\displaystyle =\frac{x^2}{2(1+x^2)}+\frac{C_1}{1+x^2}}\hfill \end{array}}$$

For $x\ge 1$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d}{dx}\left[(1+x^2)y\right]}& {\displaystyle =-x}\hfill \\ \hfill {\displaystyle (1+x^2)y}& {\displaystyle =-\frac{1}{2}{x}^2+C_2}\hfill \\ \hfill {\displaystyle y}& {\displaystyle =-\frac{x^2}{2(1+x^2)}+\frac{C_2}{1+x^2}}\hfill \end{array}}$$

The initial condition is at $x=0$, which is in the first piece: $y(0)=0\Rightarrow C_1=0$. So

$${\displaystyle y=\{\begin{array}{cc}{\displaystyle \frac{x^2}{2(1+x^2)},}\hfill & {\displaystyle 0\le x<1}\hfill \\ {\displaystyle -\frac{x^2}{2(1+x^2)}+\frac{C_2}{1+x^2},}\hfill & {\displaystyle x\ge 1}\hfill \end{array}}$$

To find $C_2$, use the requirement that $y$ is continuous, so $y$ is continuous at $x=1$.

At $x=1$, the formula on the left and the formula on the right must give the same limit:

$${\displaystyle \underset{x\to 1^{-}}{lim}y=\underset{x\to 1^{+}}{lim}y\Rightarrow \frac{1}{4}=-\frac{1}{4}+\frac{C_2}{2}\Rightarrow C_2=1}$$

With $C_2=1$, the second formula becomes ${\displaystyle -\frac{x^2}{2(1+x^2)}+\frac{1}{1+x^2}=\frac{2-x^2}{2(1+x^2)}}$:

For $0\le x<1$, the graph rises from the origin to $(1,\frac{1}{4})$. For $x\ge 1$, it decreases from $(1,\frac{1}{4})$, crosses the $x$-axis at $x=\sqrt{2}$, and levels off toward $y=-\frac{1}{2}$. The two pieces meet at $(1,\frac{1}{4})$ with a corner, so $y$ is continuous but not differentiable there.