MATH 308 Lecture 2

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Lecture Notes

Exercise 4

$\begin{array}{rl}{p}^{\prime }(t)& =0.03p(t)\\ \frac{p^{\prime }(t)}{p(t)}& =0.03\\ \ln (p(t))& =0.03t+C\\ p(t)& ={\mathrm{e}}^{0.03t+C}\\ & ={\mathrm{e}}^C{\mathrm{e}}^{0.03t}\\ & =A{\mathrm{e}}^{0.03t}\end{array}$

This is called the general solution

Given intial condition $p(0)=200$, we can find particular values for the unknown constants in the general solution:

$\begin{array}{rl}p(0)& =A{\mathrm{e}}^{0.03\cdot 0}\\ A& =200\\ p(t)& =200{\mathrm{e}}^{0.03t}\end{array}$

This unique solution is called the particular solution

In Maple, you can solve Diff Eq's as follows:

> eq1 := p'(t) = 0.03*p(t);
> dsolve(eq1, p(t));               # Gives general solution
> dsolve({eq1, p(0) = 200}, p(t)); # Gives particular Solution

Exercise 5

Show that $y=(Ax+B){\mathrm{e}}^{-x}+1$ is a solution to $y^{″}+2y^{\prime }+y=1$.

$\begin{array}{rl}{y}^{\prime }& =(A-B-Ax){\mathrm{e}}^{-x}\\ y^{″}& =-A{\mathrm{e}}^{-x}-(A-B-Ax){\mathrm{e}}^{-x}\\ (-A{\mathrm{e}}^{-x}-(A-B-Ax){\mathrm{e}}^{-x})+2((A-B-Ax){\mathrm{e}}^{-x})+((Ax+B){\mathrm{e}}^{-x}+1)& =1\\ {\mathrm{e}}^{-x}((-2A+B+2A-2B+B)+x(A-2A+A))+1& =1\\ {\mathrm{e}}^{-x}(0+x\left(0\right))+1& =1\end{array}$

Therefore, $(Ax+B){\mathrm{e}}^{-x}+1$ is a solution.

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

$\begin{array}{rl}y(0)& =(0A+B){\mathrm{e}}^0+1=3=B+1\\ y^{\prime }(0)& =(A-B-0A){\mathrm{e}}^0=1=A-B\\ B& =2\\ A& =3\end{array}$

Direction Fields

Works for first-order differential equations only!

Example: $y^{\prime }(x)=x^2-y^2$

To plot in Maple:

> with(DETools):
> DEplot(D(y)(x) = x^2-y(x)^2, y(x), x=-10..10, y=-10..10);

A solution has a horizontal tangent at $(x,y)$ if $y^{\prime }(x)=0=x^2-y^2$, so for $x=\pm y$, the direction field lines are horizontal.

Are these values minima, maxima, or inflection points? Use second derivative test

$y^{″}(x)=2x$

• For $x\ge 0$, it's a local minimum
• For $x\le 1$, it's a local maximum
• For $x=0$, we don't know

Particular solutions can be plotted in the direction fields:

> with(DETools):
> DEplot(D(y)(x) = x^2-y(x)^2, y(x), x=-10..10, y=-10..10, [[0,1], [1,-0.5], [-1,-2], [-3,0]], linecolor=blue);

Until Next Time

• Homework 1
• PRACTICE! (use suggested homework)