MATH 308 Lecture 4

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Lecture Notes

Homework Problem 6

Find value of $y_0$ for which the solution to $y^{\prime }-y=1+3\sin t$ with initial condition $y(0)=y_0$ remains finite as $t$ approaches $\mathrm{\infty }$.

General solution

$P(t)=-1$, so integrating factor is $\mu (t)={\mathrm{e}}^{-t}$.

$\begin{array}{rl}{\mathrm{e}}^{-t}{y}^{\prime }-{\mathrm{e}}^{-t}y& ={\mathrm{e}}^{-t}+3{\mathrm{e}}^{-t}\sin t\\ \frac{\mathrm{d}}{\mathrm{d}t}\left({\mathrm{e}}^{-t}y\right)& ={\mathrm{e}}^{-t}+3{\mathrm{e}}^{-t}\sin t\\ {\mathrm{e}}^{-t}y& =-{\mathrm{e}}^{-t}+3\int {\mathrm{e}}^{-t}\sin t\mathrm{d}t+C\\ y(t)& =-1-\frac{3}{2}\cos t-\frac{3}{2}\sin t+C{\mathrm{e}}^t\end{array}$

sine and cosine are bounded functions, so the function $y(t)$ will be bounded for $C{\mathrm{e}}^t$ only when $C=0$.

$y(0)=-1+\frac{3}{2}(-1)=-\frac{5}{2}$

Complex Numbers

$\underset{\text{real part}}{\underbrace{a}}+i\underset{\text{imaginary part}}{\underbrace{b}}$

Plot on coordinate plane with $(x,y)=(a,b)$; imaginary number can also be represented as magnitude $r=\sqrt{a^2+b^2}$ and angle $\theta$ made with $x$-axis:

$a+ib=r{\mathrm{e}}^{i\theta }=r\cos \theta +ir\sin \theta$

Therefore, $\sin t=\mathrm{\Im }\left({\mathrm{e}}^{it}\right)$

We can use this to avoid integration by parts, and it will come in handy for Chapter 3.

Separable Functions

Try to separate the variables to each side of the equation:

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

Separable if the RHS can be expressed as product of 2 functions: $g(x)$ and $p(y)$

Exercise 1

$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}x}& =(y^2+1)x\\ \frac{\mathrm{d}y}{y^2+1}& =x\mathrm{d}x\\ \int \frac{\mathrm{d}y}{y^2+1}& =\int x\mathrm{d}x\\ {\tan }^{-1}y& =\frac{x^2}{2}+C\\ y& =\tan (\frac{x^2}{2}+C)\end{array}$

• The solution ${\tan }^{-1}y=\frac{x^2}{2}+C$ is called the implicit solution
• The solution $y=\tan (\frac{x^2}{2}+C)$ is the explicit solution

Exercise 2

$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}x}& =x{\mathrm{e}}^{y+x^2}\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =x{\mathrm{e}}^y{\mathrm{e}}^{x^2}\\ \int \frac{\mathrm{d}y}{{\mathrm{e}}^y}& =\int x{\mathrm{e}}^{x^2}\mathrm{d}x\\ -{\mathrm{e}}^{-y}& =-\frac{1}{2}{\mathrm{e}}^{x^2}-C\\ -y& =\ln (-\frac{1}{2}{\mathrm{e}}^{x^2}-C)\\ y& =-\ln (-\frac{1}{2}{\mathrm{e}}^{x^2}-C)\end{array}$