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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'' + 2y' + y = 1} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} 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{align}}

Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (Ax + B)\mathrm{e}^{-x} + 1} is a solution.

Find a solution that satisfies the initial condition Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(0) = 3} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'(0) = 1} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} 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{align}}

Direction Fields

Works for first-order differential equations only!

Example: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y)} if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'(x) = 0 = x^2 - y^2} , so for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \pm y} , the direction field lines are horizontal.

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y''(x) = 2x}

• For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \ge 0} , it's a local minimum
• For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \le 1} , it's a local maximum
• For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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)