MATH 308 Lecture 2

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Lecture Notes

Exercise 4

${\begin{aligned}p'(t)&=0.03p(t)\\{\frac {p'(t)}{p(t)}}&=0.03\\\ln {\left(p(t)\right)}&=0.03t+C\\p(t)&=\mathrm {e} ^{0.03t+C}\\&=\mathrm {e} ^{C}\,\mathrm {e} ^{0.03t}\\&=A\,\mathrm {e} ^{0.03t}\end{aligned}}$

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:

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

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 $y''+2y'+y=1$.

${\begin{aligned}y'&=(A-B-A\,x)\mathrm {e} ^{-x}\\y''&=-A\,\mathrm {e} ^{-x}-\left(A-B-A\,x\right)\mathrm {e} ^{-x}\\\left(-A\,\mathrm {e} ^{-x}-\left(A-B-A\,x\right)\mathrm {e} ^{-x}\right)+2\left((A-B-A\,x)\mathrm {e} ^{-x}\right)+\left((A\,x+B)\mathrm {e} ^{-x}+1\right)&=1\\\mathrm {e} ^{-x}\left(\left(-2A+B+2A-2B+B\right)+x\left(A-2A+A\right)\right)+1&=1\\\mathrm {e} ^{-x}\left(0+x\left(0\right)\right)+1&=1\end{aligned}}$

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

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

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

Direction Fields

Works for first-order differential equations only!

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

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 $y'(x)=0=x^{2}-y^{2}$, 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 $x\geq 0$, it's a local minimum
• For $x\leq 1$, 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)