MATH528

MATH528Due date: 09/05/18

Homework 02 Solution

1Exercises

Exercise. PS1.4.1

Solution. $2 x y dx + x^{2} d y$ is an exact differential

In[1]:=

df = 2 x y dx + x^2 dy

$2 dx x y + dy x^{2}$

In[4]:=

{P[x_,y_],Q[x_,y_]} = {Coefficient[df,dx,1],Coefficient[df,dy,1]}

${2 x y, x^{2}}$

In[5]:=

ExactDF = D[P[x,y],y] == D[Q[x,y],x]

$True$

In[6]:=

Exercise. PS1.4.2

Solution. $x^{3} d x + y^{3} d y$ is an exact differential

In[6]:=

df = x^3 dx + y^3 dy;

{P[x_,y_],Q[x_,y_]} = {Coefficient[df,dx,1],Coefficient[df,dy,1]};

ExactDF = D[P[x,y],y] == D[Q[x,y],x];

{P[x,y],Q[x,y],ExactDF}

${x^{3}, y^{3}, True}$

In[7]:=

Exercise. PS1.4.3

Solution. $\sin x \cos y d x + \cos x \sin y d y$ is an exact differential

In[7]:=

df = Sin[x] Cos[y] dx + Cos[x] Sin[y] dy;

{P[x_,y_],Q[x_,y_]} = {Coefficient[df,dx,1],Coefficient[df,dy,1]};

ExactDF = D[P[x,y],y] == D[Q[x,y],x];

{P[x,y],Q[x,y],ExactDF}

${\sin (x) \cos (y), \cos (x) \sin (y), True}$

In[8]:=

Exercise. PS1.4.4

Solution. $e^{3 θ} (d r + 3 r d θ)$ is an exact differential

In[8]:=

df = Exp[3y](dx + 3x dy);

{P[x_,y_],Q[x_,y_]} = {Coefficient[df,dx,1],Coefficient[df,dy,1]};

ExactDF = D[P[x,y],y] == D[Q[x,y],x];

{P[x,y],Q[x,y],ExactDF} /. {x->r, y->theta}

${e^{3 θ}, 3 r e^{3 θ}, True}$

In[9]:=

Exercise. PS1.5.3

Solution. $y^{'} - y = 5.2$ is a linear, first-order ODE of form $y^{'} + p (x) y = r (x)$ , with $p (x) = - 1$ , $r (x) = 5.2$ , and general solution

y (x) = e^{- h (x)} (\int e^{h (x)} r (x) d x + c), h = \int p (x) d x,

which gives

h (x) = - x

y (x) = e^{x} (5.2 \int e^{- x} d x + c) = - 5.2 + c e^{x} .

Verify:

In[23]:=

ODE = y'[x] - y[x] == 5.2;

sol[t] = y[x] /. DSolve[ODE,y[x],x][[1,1]]

$c_{1} e^{1 . x} - 5.2$

In[25]:=

Integrate[-4 x Exp[-2x],x]

$- 4 e^{- 2 x} (- \frac{x}{2} - \frac{1}{4})$

In[26]:=

Exercise. PS1.5.4

Solution. $y^{'} = 2 y - 4 x$ , linear ODE with $p (x) = - 2$ , $r (x) = - 4 x$ , $h (x) = - 2 x$

y (x) = e^{2 x} (- 4 \int x e^{- 2 x} d x + c) = e^{2 x} (2 x e^{- 2 x} - \frac{1}{2} \int e^{- 2 x} d x + c) = e^{2 x} (2 x e^{- 2 x} + e^{- 2 x} + c),

by integration by parts. Verify:

In[24]:=

ODE = y'[x] == 2y[x] -4x;

sol[t] = y[x] /. DSolve[ODE,y[x],x][[1,1]]

$c_{1} e^{2 x} - 4 (- \frac{x}{2} - \frac{1}{4})$

In[25]:=

Exercise. PS1.5.5

Solution. $y^{'} + k y = e^{- k x}$ , linear ODE with $p (x) = k$ , $r (x) = e^{- k x}$

In[27]:=

ODE = y'[x] + k y[x] == Exp[-k x];

sol[t] = y[x] /. DSolve[ODE,y[x],x][[1,1]]

$c_{1} e^{- k x} + x e^{- k x}$

In[28]:=

Exercise. PS1.5.6

Solution. $y^{'} + 2 y = 4 \cos 2 x$ , $y (π / 4) = 3$ , is a linear ODE with $p (x) = 2$ , $r (x) = 4 \cos 2 x$

In[29]:=

ODE = y'[x] + 2 y[x] == 4 Cos[2x];

iCond = y[Pi/4]==3;

sol[t] = Expand[y[x] /. DSolve[{ODE,iCond},y[x],x][[1,1]]]

$2 e^{\frac{π}{2} - 2 x} + \sin (2 x) + \cos (2 x)$

In[30]:=

2Problems

Problem. PS1.5.31 Newton's law of cooling

Solution. Assume cooling rate is proportional to temperature difference $T (t) - T_{r}$ , with time denoted by $t$ and $T_{r}$ the room temperature. The problem can be formulated as

{\begin{cases} \frac{d T}{d t} = - r (T - T_{r}) \\ T (0) = 300 \end{cases} .

and also written

{\begin{cases} \frac{d Y}{d t} = - r Y \\ Y (0) = 240 \end{cases} .

with $Y = T (t) - T_{r}$ , with solution $Y (t) = 240 e^{- r t}$ , or $T (t) = 240 e^{- r t} + 60$

In[20]:=

ODE = T'[t] == -r (T[t]-60);

sol[t_] = Expand[T[t] /. DSolve[{ODE,T[0]==300},T[t],t][[1,1]]]

$240 e^{- r t} + 60$

In[21]:=

Find the decay rate $r$ by imposing the condition $T (10) = 200$

In[21]:=

rsol = FindRoot[sol[10.] == 200., {r,0.1}][[1]]

$r \to 0.0538997$

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Find time at which cake temperature reaches 61 degrees

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tsol = FindRoot[sol[t]==61 /. rsol,{t,20}][[1]]

$t \to 101.682$

In[23]:=

Verify the solution

In[14]:=

sol[t] /. {rsol,tsol}

$61 .$

Problem. PS1.5.32 Heating and cooling of a building

Solution. Define the ODE and solve

In[1]:=

Ta[t_] = A - c Cos[2 Pi t/24]

$A - c \cos (\frac{π t}{12})$

In[2]:=

ODE = T'[t] == Subscript[k,1] (T[t]-Ta[t]) + Subscript[k,2] (T[t] - Subscript[T,w]) + P

$T^{'} (t) = k_{1} (- A + c \cos (\frac{π t}{12}) + T (t)) + k_{2} (T (t) - T_{w}) + P$

In[3]:=

The above is a linear ODE of form $T^{'} + p (t) T = r (t)$ , with

p (t) = - (k_{1} + k_{2}), r (t) = k_{1} (- A + c \cos (\frac{π t}{12})) - k_{2} T_{ω} + P

The general solution is

T (t) = e^{- h (t)} (\int e^{h (t)} r (t) d t + B), h = \int p (t) d t = - (k_{1} + k_{2}) t .

In[3]:=

h[t_]=-(Subscript[k,1]+Subscript[k,2])t

$(k_{1} + k_{2}) (- t)$

In[4]:=

r[t_]=Subscript[k,1]Ta[t]-Subscript[k,2] Subscript[T,w]+P

$k_{1} (A - c \cos (\frac{π t}{12})) - k_{2} T_{w} + P$

In[5]:=

T[t_] = FullSimplify[Exp[-h[t]](Integrate[Exp[h[t]] r[t],t]+B)]

$- \frac{A k_{1} - k_{2} T_{w} + P}{k_{1} + k_{2}} + B e^{(k_{1} + k_{2}) t} - \frac{12 π c k_{1} \sin (\frac{π t}{12})}{144 {(k_{1} + k_{2})}^{2} + π^{2}} + \frac{144 c k_{1} (k_{1} + k_{2}) \cos (\frac{π t}{12})}{144 {(k_{1} + k_{2})}^{2} + π^{2}}$

In[6]:=

The above terms are, in order:

temperature balance between ambient, wanted, occupant temperatures
decay of initial temperature
4. variation due to change in ambient temperature

A more detailed investigation is presented in mLab03.

Problem. PS1.5.33 Drug injection

Solution. Let $d (t)$ be the amount of drug in the bloodstream. The model is written as

d^{'} = A - r d

with $A$ the injection rate and $r$ the removal rate per unit of drug in bloodstream. This is a linear ODE

In[6]:=

ODE = d'[t] == A - r d[t];

sol[t] = d[t] /. DSolve[ODE,d[t],t][[1,1]]

$\frac{A}{r} + c_{1} e^{- r t}$

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Problem. PS1.5.35 Lake Erie

Solution. Volume $V = 450 {km}^{3}$ , flow rate $Q = 175 {km}^{3} / year$ (in and out). Let $p (t)$ denote pollution concentration, with $p (0) = 0.04$ . The inflow pollution is $p_{1} = p (0) / 4 = 0.01$ . The model is

p^{'} = \frac{Q}{V} (p_{1} - p), p (0) = 0.04 .

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V=450; Q=175; p0=0.04; p1=p0/4; ODE = p'[t] == Q/V (p1-p[t]); iCond = p[0]==p0;

sol[t] = p[t] /. DSolve[{ODE,iCond},p[t],t][[1,1]]

$0.01 e^{- 0.388889 t} (3 . + 1 . e^{0.388889 t})$

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FindRoot[sol[t]==p0/2,{t,1}]

${t \to 2.825}$

In[10]:=

2.825*175

$494.375$

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3Projects

3.1PS2.1.16

The sum $y = y_{1} + y_{2}$ is a solution of $y^{'} + p y = r_{1} + r_{2}$ .

Proof. Compute $y^{'} + p y = y_{1}^{'} + y_{2}^{'} + p (y_{1} + y_{2}) = (y_{1}^{'} + p y_{1}) + (y_{2}^{'} + p y_{2}) = r_{1} + r_{2}$