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Intro (Newton example):
Newton example where we find the equations to describe a falling object using differential equations (DE’s)
We know $F=ma$
$F=m\frac{dv}{dt}=mg-kv$ <- we account for air resistance here. We can approximate the force of air resistance is …">
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Intro (Newton example): Newton example where we find the equations to describe a falling object using differential equations (DE’s) We know $F=ma$ $F=m\frac{dv}{dt}=mg-kv$ <- we account for air resistance here. We can approximate the force of air resistance is proportional to the speed times a constant. We can rearrange and solve it as it is a separable DE: $\frac{dv}{mg-kv}=\frac{dt}{m}$ integrating both sides: $\int \frac{dv}{mg-kv}=\frac{t}{m}+C$ let $u=mg-kv \quad du=-kdv$ $\int \frac{dv}{mg-kv}=\int \frac{du}{-k*u}=\frac{1}{-k}\ln\mid mg-kv\mid=\frac{t}{m}+C$ Very cool, but I want the velocity as a function of time, isolate v $\ln\mid mg-kv\mid=-\frac{kt}{m}+C$ $\mid mg-kv\mid=e^{\frac{-kt}{m}+C}=e^{\frac{-kt}{m}}e^C$ $e^C$ is a + constant, the absolute value will multiply the inside expression by -1 when the inside is negative, so we can replace the $e^C$ constant with an arbitrary constant A that can be + or - $mg-kv=Ae^{\frac{-kt}{m}}$ so, the general solution is $v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$" />
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Intro (Newton example): Newton example where we find the equations to describe a falling object using differential equations (DE’s) We know $F=ma$ $F=m\frac{dv}{dt}=mg-kv$ <- we account for air resistance here. We can approximate the force of air resistance is proportional to the speed times a constant. We can rearrange and solve it as it is a separable DE: $\frac{dv}{mg-kv}=\frac{dt}{m}$ integrating both sides: $\int \frac{dv}{mg-kv}=\frac{t}{m}+C$ let $u=mg-kv \quad du=-kdv$ $\int \frac{dv}{mg-kv}=\int \frac{du}{-k*u}=\frac{1}{-k}\ln\mid mg-kv\mid=\frac{t}{m}+C$ Very cool, but I want the velocity as a function of time, isolate v $\ln\mid mg-kv\mid=-\frac{kt}{m}+C$ $\mid mg-kv\mid=e^{\frac{-kt}{m}+C}=e^{\frac{-kt}{m}}e^C$ $e^C$ is a + constant, the absolute value will multiply the inside expression by -1 when the inside is negative, so we can replace the $e^C$ constant with an arbitrary constant A that can be + or - $mg-kv=Ae^{\frac{-kt}{m}}$ so, the general solution is $v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$"/>
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<p>#start of lecture 1</p>
<h3 id="intro-newton-example">Intro (Newton example):</h3>
<p>Newton example where we find the equations to describe a falling object using differential equations (DE’s)
We know $F=ma$
$F=m\frac{dv}{dt}=mg-kv$ <svg width="11px" height="10px" viewBox="0 0 11 10" version="1.1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink"> <g id="left-arrow" transform="translate(5.500000, 5.000000) scale(-1, 1) translate(-5.500000, -5.000000) "> <path d="M1.77635684e-14,5 L9,5" id="rod" stroke="#000000" stroke-width="2"></path> <path d="M11,5 L6,0.5 L6,9.5 L11,5 Z" id="point" fill="#000000"></path></g></svg> we account for air resistance here. We can approximate the force of air resistance is proportional to the speed times a constant.
We can rearrange and solve it as it is a separable DE:
$\frac{dv}{mg-kv}=\frac{dt}{m}$
integrating both sides:
$\int \frac{dv}{mg-kv}=\frac{t}{m}+C$
let $u=mg-kv \quad du=-kdv$
$\int \frac{dv}{mg-kv}=\int \frac{du}{-k*u}=\frac{1}{-k}\ln\mid mg-kv\mid=\frac{t}{m}+C$
Very cool, but I want the velocity as a function of time, isolate v
$\ln\mid mg-kv\mid=-\frac{kt}{m}+C$
$\mid mg-kv\mid=e^{\frac{-kt}{m}+C}=e^{\frac{-kt}{m}}e^C$
$e^C$ is a + constant, the absolute value will multiply the inside expression by -1 when the inside is negative, so we can replace the $e^C$ constant with an arbitrary constant A that can be + or -
$mg-kv=Ae^{\frac{-kt}{m}}$
so, the general solution is $v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$</p>
<h3 id="separable-de">Separable DE:</h3>
<pre><code>$\frac{dy}{dx}=f(y)g(x) \rightarrow \frac{dy}{f(y)}=g(x)dx\quad where\quad f(y)\ne0$
(I'm calling this <a class="hashtag" onclick="focusTag(this)">de_s_type1)</a>
</code></pre>
<p>ex: $\frac{dy}{dt}=\frac{1-t^2}{y^2}$
$y^2dy=dt(1-t^2)$
integrating both sides yields:
$\frac{y^3}{3}=t-\frac{t^3}{3}+C$
$y=(3t-t^3+C)^\frac{1}{3}$</p>
<h3 id="initial-value-problem-ivp">Initial value problem (IVP):</h3>
<pre><code>A Differential equation with provided initial conditions.
</code></pre>
<p><a class="hashtag" onclick="focusTag(this)">ex</a>
<a class="hashtag" onclick="focusTag(this)">IVP</a> <a class="hashtag" onclick="focusTag(this)">de_s_type1</a>
ex: $\frac{dy}{dx}=2x\cos^2(y), \quad y(0)=\frac{\pi}{4}$
$\frac{dy}{\cos^2(y)}=2xdx$
integrate both sides yields:
$\int \frac{dy}{\cos^2(y)}=\tan(y)+C=x^2$
plug in $y(0)=\frac{\pi}{4}$
$\tan\left( \frac{\pi}{4} \right)+C=0$
$1+C=0$
$C=-1$
So, the answer is: $y=\arctan(x^2+1)$</p>
<p><a class="hashtag" onclick="focusTag(this)">end</a> of Lecture 1</p>
<h3>Referenced in</h3>
<ul>
<li>No backlinks found</li>
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