MATH 308 Lecture 5

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Lecture Notes

${\begin{aligned}y\,\mathrm {e} ^{-x}\,y'+x&=0\\y\,y'=-x\,\mathrm {e} ^{x}\\{\frac {1}{2}}\,y^{2}&=-\int x\,\mathrm {e} ^{x}\,\mathrm {d} x\\&=-x\,\mathrm {e} ^{x}+\int \mathrm {e} ^{x}\,\mathrm {d} x\\&=-x\,\mathrm {e} ^{x}+\mathrm {e} ^{x}+C\\&=\mathrm {e} ^{x}\left(1-x\right)+C\\y^{2}&=2\mathrm {e} ^{x}\left(1-x\right)+C\\y&=\pm {\sqrt {2\mathrm {e} ^{x}\left(1-x\right)+C}}\\\end{aligned}}$

If $y(0)=1$ then we take the positive square root

${\begin{aligned}1&={\sqrt {2+C}}\\1&=2+C\\C&=-1\end{aligned}}$

Therefore the particular solution is

$y=\pm {\sqrt {2\mathrm {e} ^{x}\left(1-x\right)-1}}$

As $x$ approaches positive infinity, the square root does not exist.

Exercise 1

Filling a container with a leak:

• Max capacity: 500 gal
• At Time $t=0$, there is 200 gal of water in the container with 100 lb of salt dissolved
• Salt water flows in at a rate of 3 gal/min with 1 lb/gal of salt dissolved
• The fluid leaks from the container at a rate of 2 gal/min

Find the amount of salt at the time the container is about to overflow.

First let's find an equation to represent the volume since we will need it to calculate the concentration of salt in the container.

${\begin{aligned}V(t)&=(200{\text{gal}})+(t{\text{min}})\cdot ((3{\text{gal/min}})-(2{\text{gal/min}}))\\&=200+t\end{aligned}}$

Now let's put together a formula to represent the rate of change of salt in the container. Let $A(t)$ represent the amount of salt in lbs at time $t$ in minutes.

${\begin{aligned}{\frac {\mathrm {d} A}{\mathrm {d} t}}&=(1{\text{lb/gal}})\cdot (3{\text{gal/min}})-(2{\text{gal/min}})\cdot {\frac {(A(t){\text{lb}})}{(V(t){\text{gal}})}}\\&=3-{\frac {2A(t)}{200+t}}\end{aligned}}$

The equation is a 2^nd order linear differential equation, so we can use the integrating factor method to solve it

${\begin{aligned}{\frac {\mathrm {d} A}{\mathrm {d} t}}+\left({\frac {2}{200+t}}\right)\,A(t)&=3\\P(x)&={\frac {2}{200+t}}\\\mu (x)&=\mathrm {e} ^{2\ln {(200+t)}}=(200+t)^{2}\\\end{aligned}}$

Now we can solve the equation:

${\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}\left((200+t)^{2}\,A(t)\right)&=3(200+t)^{2}\\(200+t)^{2}\,A(t)&=(200+t)^{3}+C\\A(t)&=200+t+{\frac {C}{(200+t)^{2}}}\end{aligned}}$

Exercise 2

A young person with no initial capital invests $k$ dollars per year at an annual return $r$. Assume that the investments are made continuously and that the return is compounded continuously.

Determine the sum $S(t)$ accumulated at time $t$

${\frac {\mathrm {d} S(t)}{\mathrm {d} t}}=k+r\,S(t)$

Since there is nothing in the bank at the start, our initial condition is $S(0)=0$.

We can use either the separable method or the integrating factor method to solve:

${\begin{aligned}\int {\frac {\mathrm {d} S(t)}{k+r\,S(t)}}&=\int \mathrm {d} t\\{\frac {\ln {(k+r\,S(t))}}{r}}&=t+C\\k+r\,S(t)&=\exp {(r\,t+r\,C)}\\S(t)&={\frac {\exp {(r\,t+r\,C)}-k}{r}}\end{aligned}}$

Exercise 3

A ball with mass 0.15 kg is thrown upward with an initial velocity of 20 m/s from the roof of a building 30 m high. There is a force due to air resistance of ${\tfrac {\left|v\right|}{30}}$, where the velocity $v$ is measured in m/s.

$m\,{\frac {\mathrm {d} v}{\mathrm {d} t}}=-m\,g-{\frac {v}{30}}$

What is acceleration?

${\frac {\mathrm {d} v}{\mathrm {d} t}}={\frac {\mathrm {d} v}{\mathrm {d} x}}\,{\frac {\mathrm {d} x}{\mathrm {d} t}}$

So $m\,{\frac {\mathrm {d} v}{\mathrm {d} t}}\cdot v=-mg-{\frac {v}{30}}$