MATH 308 Lecture 10

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Lecture Notes

Chapter 3

Solving second order linear differential equations

Section 3.1–3.4: Homogeneous with constant coefficients

$P(x)y^{″}+Q(x)y^{\prime }+R(x)y=G(x)$

When $G(x)=0$, equation is homogeneous When $P(x)$, $Q(x)$, and $R(x)$ do not depend on $x$, the equation has constant coefficients

Examples:

Equation	Linear	Homogeneous	Constant Coefficients
$x{y}^{″}+3y^{\prime }+y\sin x={\mathrm{e}}^x$	Yes	No	No
$y^{″}+x{y}^{\prime }=(y-x)x$	Yes	No	No
$y^{″}+xy{y}^{\prime }=xy$	No	--	--
$y^{″}=y\sin x$	Yes	Yes	No
$y^{″}+5y^{\prime }+3y=\sin x$	Yes	No	Yes
$y^{″}-2y^{\prime }=5\sqrt{y}$	No	--	--

Theorem

Let $y_1$ and $y_2$ be two solutions to a second order homogeneous linear differential equation. Any linear combination $c_1{y}_1+c_2{y}_2$, for any real $c_1$ and $c_2$ is a solution to the differential equation.

Exercise 2

Given that $y_1(x)={\mathrm{e}}^{3x}$ and $y_2(x)={\mathrm{e}}^{2x}$ are solutions to the homogeneous equation

$y^{″}-5y^{\prime }+6y=0$

Show that every linear combination $c_1{y}_1+c_2{y}_2$ for any $c_1,c_2\in \mathbb{\mathbb{Z}}$ is a solution to the homogeneous problem.

$\begin{array}{rlr}f(x)& =c_1{\mathrm{e}}^{3x}+c_2{\mathrm{e}}^{2x}{f}^{\prime }(x)& =c_1{\mathrm{e}}^{3x}+c_2{\mathrm{e}}^{2x}\end{array}$

$\begin{array}{rl}f(x)& =c_1{y}_1+c_2{y}_2\\ f^{\prime }(x)& =c_1{y_1}^{\prime }+c_2{y_2}^{\prime }\\ f^{″}(x)& =c_1{y_1}^{″}+c_2{y_2}^{″}\\ \end{array}$

$\begin{array}{rl}{c}_1{y_1}^{″}+c_2{y_2}^{″}-5(c_1{y_1}^{\prime }+c_2{y_2}^{\prime })+6(c_1{y}_1+c_2{y}_2)& =0\\ c_1({y_1}^{″}-5{y_1}^{\prime }+6y_1)+c_2({y_2}^{″}-5{y_2}^{\prime }+6y_2)& =0\\ \end{array}$

• y_1 is a solution: ${y_1}^{″}-5{y_1}^{\prime }+6y_1=0$
• y_2 is a solution: ${y_2}^{″}-5{y_2}^{\prime }+6y_2=0$

Therefore $f$ is a solution.

Find a solution that satisfies the initial condition $y(0)=3$ and $y^{\prime }(0)=-1$.

Find $f$ in the form $c_1{\mathrm{e}}^{3x}+c_2{\mathrm{e}}^{2x}$

$\begin{array}{rl}f(x)& =c_1{\mathrm{e}}^{3x}+c_2{\mathrm{e}}^{2x}\\ f^{\prime }(x)& =3c_1{\mathrm{e}}^{3x}+2c_2{\mathrm{e}}^{2x}\\ f(0)& =c_1+c_2=3\\ f^{\prime }(0)& =3c_1+2c_2=-1\\ c_1& =-7\\ c_2& =10\end{array}$

Therefore, the solution is $-7{\mathrm{e}}^{3x}+10{\mathrm{e}}^{2x}$

Exercise 3

Find the general solution to the differential equation $y^{″}-7y^{\prime }+12y=0$. Look for a solution of the form $y={\mathrm{e}}^{rx}$.

$\begin{array}{rl}y& ={\mathrm{e}}^{rx}\\ y^{\prime }& =r{\mathrm{e}}^{rx}\\ y^{″}& =r^2{\mathrm{e}}^{rx}\end{array}$

So we have

$\begin{array}{rl}{r}^2{\mathrm{e}}^{rx}-7r{\mathrm{e}}^{rx}+12{\mathrm{e}}^{rx}& =0\\ r^2-7r+12& =0\end{array}$

This is called the characteristic equation.

Solving for $r$ gives $\{3,4\}$, so the following formulae are solutions:

• $y_1={\mathrm{e}}^{3x}$
• $y_2={\mathrm{e}}^{4x}$

And any linear combination $y=c_1{\mathrm{e}}^{3x}+c_2{\mathrm{e}}^{4x}$ is the general solution to the differential equation.

Characteristic Equation

Replace each derivative with a coefficient of the same degree.

For example, $y^{″}-3y^{\prime }-10y=0$ becomes $r^2-3r-10=0$.

Solving the characteristic gives coefficients for the exponential.

Exercise 4

Find the general solutino to the differential solution $y^{″}+4y=0$.

Characteristic equation is $r^2+4=0$, and roots are $r=\pm 2i$.

General complex valued solution is $y=c_1{\mathrm{e}}^{2ix}+c_2{\mathrm{e}}^{-2ix}$.

Recall the Euler Formula: ${\mathrm{e}}^{i\theta }=\cos \theta +i\sin \theta$

The solution takes the form

$\begin{array}{rl}{y}_1& =\cos 2x+i\sin 2x\\ y_2& =\cos -2x+i\sin -2x=\cos 2x-i\sin 2x\\ \end{array}$

Recombining we get

• Let $c_1=c_2=\frac{1}{2}$, then $\frac{y_1+y_2}{2}=\cos 2x$
• Let $c_1=c_2=\frac{1}{2i}$, then $\frac{y_1+y_2}{2}=\sin 2x$.

So we can express the solution as a linear combination of these two real functions:

$y=c_1\cos 2x+c_2\sin 2x$