MATH 308 Lecture 4

« previous | Wednesday, January 23, 2013 | next »

Lecture Notes

Homework Problem 6

Find value of $y_{0}$ for which the solution to $y'-y=1+3\sin {t}$ with initial condition $y(0)=y_{0}$ remains finite as $t$ approaches $\infty$.

General solution

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

${\begin{aligned}\mathrm {e} ^{-t}\,y'-\mathrm {e} ^{-t}\,y&=\mathrm {e} ^{-t}+3\mathrm {e} ^{-t}\sin {t}\\{\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathrm {e} ^{-t}\,y\right)&=\mathrm {e} ^{-t}+3\mathrm {e} ^{-t}\sin {t}\\\mathrm {e} ^{-t}\,y&=-\mathrm {e} ^{-t}+3\int \mathrm {e} ^{-t}\,\sin {t}\,\mathrm {d} t+C\\y(t)&=-1-{\frac {3}{2}}\cos {t}-{\frac {3}{2}}\sin {t}+C\,\mathrm {e} ^{t}\end{aligned}}$

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

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

Complex Numbers

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

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:

$a+i\,b=r\mathrm {e} ^{i\,\theta }=r\cos {\theta }+i\,r\sin {\theta }$

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

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

${\begin{aligned}{\frac {\mathrm {d} y}{\mathrm {d} x}}&=\left(y^{2}+1\right)\,x\\{\frac {\mathrm {d} y}{y^{2}+1}}&=x\,\mathrm {d} x\\\int {\frac {\mathrm {d} y}{y^{2}+1}}&=\int x\,\mathrm {d} x\\\tan ^{-1}{y}&={\frac {x^{2}}{2}}+C\\y&=\tan {\left({\frac {x^{2}}{2}}+C\right)}\end{aligned}}$

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

Exercise 2

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