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

$P(x)y''+Q(x)y'+R(x)y=G(x)$

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
$x\,y''+3y'+y\,\sin {x}=\mathrm {e} ^{x}$	Yes	No	No
$y''+x\,y'=(y-x)x$	Yes	No	No
$y''+x\,y\,y'=x\,y$	No	--	--
$y''=y\,\sin {x}$	Yes	Yes	No
$y''+5y'+3y=\sin {x}$	Yes	No	Yes
$y''-2y'=5{\sqrt {y}}$	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

$y''-5y'+6y=0$

Show that every linear combination $c_{1}\,y_{1}+c_{2}\,y_{2}$ for any $c_{1},c_{2}\in \mathbb {Z}$ is a solution to the homogeneous problem.

${\begin{aligned}f(x)&=c_{1}\,\mathrm {e} ^{3x}+c_{2}\,\mathrm {e} ^{2x}f'(x)&=c_{1}\,\mathrm {e} ^{3x}+c_{2}\,\mathrm {e} ^{2x}\end{aligned}}$

${\begin{aligned}f(x)&=c_{1}\,y_{1}+c_{2}\,y_{2}\\f'(x)&=c_{1}\,y_{1}'+c_{2}\,y_{2}'\\f''(x)&=c_{1}\,y_{1}''+c_{2}\,y_{2}''\\\end{aligned}}$

${\begin{aligned}c_{1}\,y_{1}''+c_{2}\,y_{2}''-5(c_{1}\,y_{1}'+c_{2}\,y_{2}')+6(c_{1}\,y_{1}+c_{2}\,y_{2})&=0\\c_{1}(y_{1}''-5y_{1}'+6y_{1})+c_{2}(y_{2}''-5y_{2}'+6y_{2})&=0\\\end{aligned}}$

• y_1 is a solution: $y_{1}''-5y_{1}'+6y_{1}=0$
• y_2 is a solution: $y_{2}''-5y_{2}'+6y_{2}=0$

Therefore $f$ is a solution.

Find a solution that satisfies the initial condition $y(0)=3$ and $y'(0)=-1$.

Find $f$ in the form $c_{1}\,\mathrm {e} ^{3x}+c_{2}\,\mathrm {e} ^{2x}$

${\begin{aligned}f(x)&=c_{1}\,\mathrm {e} ^{3x}+c_{2}\,\mathrm {e} ^{2x}\\f'(x)&=3c_{1}\,\mathrm {e} ^{3x}+2c_{2}\,\mathrm {e} ^{2x}\\f(0)&=c_{1}+c_{2}=3\\f'(0)&=3c_{1}+2c_{2}=-1\\c_{1}&=-7\\c_{2}&=10\end{aligned}}$

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

Exercise 3

Find the general solution to the differential equation $y''-7y'+12y=0$. Look for a solution of the form $y=\mathrm {e} ^{r\,x}$.

${\begin{aligned}y&=\mathrm {e} ^{rx}\\y'&=r\,\mathrm {e} ^{rx}\\y''&=r^{2}\,\mathrm {e} ^{rx}\end{aligned}}$

So we have

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

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, $y''-3y'-10y=0$ 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 $y=c_{1}\,\mathrm {e} ^{2i\,x}+c_{2}\,\mathrm {e} ^{-2i\,x}$.

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

The solution takes the form

${\begin{aligned}y_{1}&=\cos {2x}+i\,\sin {2x}\\y_{2}&=\cos {-2x}+i\,\sin {-2x}=\cos {2x}-i\,\sin {2x}\\\end{aligned}}$

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:

$y=c_{1}\,\cos {2x}+c_{2}\,\sin {2x}$