Tag Archives: Ordinary Differential Equations

Elementary Differential Equations by Dela Fuente, Feliciano and Uy Physical Application 2: Exponential Growth and Decay

A certain radioactive material follows the law of exponential change and has a half life of 38 hours. Find how long it takes for 90% of the radioactivity to be dissipated.

Solution:

Use the formula:

S=Ce^{-kt}

First, find the constant of proportionality. In the problem, after 38 hours, half of the radioactivity has been dissipated and a half has been retained. So we can assume that S = 0.5So when t = 38 hrs and C = So.

\left(0.5\right)So=\left(So\right)e^{-k\left(38\right)}

And then solve for k:

k=-0.018241

And then substitute k to the formula:

S=Ce^{-0.018241\left(t\right)}

Now we can solve for the time(t). According to the problem, 90% of the radioactivity is dissipated, so 10% is retained. So we can assume that S = 0.1So and change C = So.

\left(0.1\right)So=\left(So\right)e^{-0.018241\left(t\right)}

And then solve for time(t):

t\:=\:126.23\:hrs
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Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 10 Problem 3 — Applications of Ordinary First-Ordered Differential Equations

A tank contains 400 liters of brine. Twelve liters of brine, each containing 2.5 N of dissolved salt, enter the tank per minute, and the mixture, assumed uniform leaves at the rate of 8 liters per min. If the concentration is to be 2 N/litre at the end of one hour, how many newtons of salt should there be present in the tank originally?

Solution:

Consider the following illustration

\frac{dS}{dt}=\left(\frac{dS}{dt}\right)_{en}-\left(\frac{dS}{dt}\right)_{es}

Using,

V_{brine}+\left(rate\:of\:brine\:out\right)t\\\frac{dS}{dt}=\frac{8L}{M}\left(\frac{S}{400+4t}\right)\\=\frac{8S}{\left(400+4t\right)}\\\frac{dS}{dt}=\frac{2S}{\left(100+t\right)}

Using the general solution:

\frac{dS}{dt}=30-\frac{2S}{\left(100+t\right)}\\\frac{dS}{dt}+\frac{2S}{\left(100+t\right)}=30

To solve we will use First Order Linear Differential Equation (FOLDE) where:

P_{\left(t\right)}=\frac{2}{\left(100+t\right)}\:,\:Q_{\left(t\right)}=30

Solve for the integrating factor using the formula:

\sigma =e^{\int \:P_{\left(t\right)}dt}

Apply,

\sigma =e^{\int \:\frac{2}{100+t}dt}\\\sigma =e^{2ln\left(100+t\right)}\\\sigma \:=e^{ln\left(100+t\right)^2}\\\sigma \:=\left(100+t\right)^2

Substitute the given value to the formula:

S\sigma =\int \:\sigma Q\left(t\right)dt+C

Apply,

S\left(100+t\right)^2=\int \:\left(100+t\right)^230dt+C\\S\left(100+t\right)^2=30\int \:\left(100+t\right)^2dt+C\\S\left(100+t\right)^2=30\:\frac{\left(100+t\right)^{^3}}{3}dt+C\\S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C\rightarrow eqn.1

Evaluate C; @t=1hr

Convert 1hr to minutes, where 1hr is simply 60 minutes.

S\left(100+60\right)\:2\:=10\left(100+60\right)\:^3\\C=\frac{2N}{L}\:;\:C=\frac{S}{\left(400+4t\right)}

Get the value of S using the equation:

\:C=\frac{S}{\left(400+4t\right)}

Isolate S,

S=C\left(400+4t\right);\:C=2,\:t=60\\S=2\left(400+4\left(60\right)\right)\\S=1280N

Get the value of C using Eqn.1

S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C; S=1280 , t=60\\
1280\left(100+60\right)^2=10\left(100+60\right)^{^3}+C\\32768000=40960000+C\\32768000-40960000=C\\ C=-8192000

With the presence of the value of C we will now have our working equation:

S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000

Using the given working equation, solve for the value of S @ t=0

S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000; t=0\\S\left(100+0\right)^2=10\left(100+0\right)^{^3}-8192000\\\frac{S\left(1000\right)^2}{1000}=\frac{1808000}{1000}\\S=180.8 N
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 10 Problem 2 — Applications of Ordinary First-Ordered Differential Equations

Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.

Solution:

Plot points on the curve, 

A(x_{1},y_{1})

We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.

y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)

As the normal intersects the x-axis, y = 0

Substituting to the previous equation, we get

\begin{align*}
-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right)\\
-my_{1}&=-1\left(x-x_{1}\right)\\
-my_{1}&=-x+x_{1}\\
x&=x_{1}+my_{1}
\end{align*}

By using distance formula, from the origin (0,0), to point (x1+y1) = from intersection to (x1+y1) 

\begin{align*}
\sqrt{\left(x_{1}-0\right)^2+\left(y_{1}-0\right)^2}&=\sqrt{\left(x_{1}+my_{1}-x_{1}\right)^2+\left(0-y_{1}\right)^2}\\
\sqrt{x_{1}^2+y_{1}^2}&=\sqrt{m^2y_{1}^2+y_{1}^2}\\
x_{1}^2+y_{1}^2&=m^2y_{1}^2+y_{1}^2\\
x_{1}^2&=m^2y_{1}^2\\
x_{1}&=m_{1}y_{1}\:\:\:\:;m=\frac{dy}{dx}\\
x_{1}&=\frac{dy}{dx}y_{1}
\end{align*}

Change x1 and y1 to x and y,

\begin{align*}
x&=y\frac{dy}{dx}\\
xdx&=ydy
\end{align*}

By integrating,

\begin{align*}

\int \:ydx&=\int \:xdy\\
\frac{y^2}{2}&=\frac{x^2}{2}+C\\
y^2&={x^2}+2C\\
\end{align*}

We get,

y^2-x^2-=2C
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 1 – Special Second-Ordered Differential Equations

Find the General Solution

\frac{d}{dx}\left(\frac{dy}{dx}\right)=\:6x\:+\:3

Solution:

\frac{d}{dx}\left(\frac{dy}{dx}\right)=\:6x\:+\:3 \\ 

 solve\; the\; equation\; using\; case\; 1,\\

let\; u=\frac{dy}{dx} \\
\int \:\frac{du}{dx}\:=\:\int \:\left(6x\:+\:3\right)\\

using\;  separation\; of\; variable\; divide\;   both\;  sides\;  by\;  dx,\\

\int \:du\:=\:\int \:\left(6x\:+\:3\right)dx\\

by\; integrating\; using\; the\; sum\; rule:\\
we\; get,\\

u=3x^2\:+\:3x\:+\:C_1\\

substitute\; the\; value\; of\; u=\frac{dy}{dx} \\

\frac{dy}{dx}=3x^2\:+\:3x\:+\:C_1\\

using\;  separation\; of\; variable:\\

\int \:dy = \int \:\left(3x^2+3x\:+\:C_1\right)dx\\

apply\; the\; sum\; rule:\\

\int \:dy=\int \:3x^2dx+\int  \:3xdx+ \int \:C_1dx\\

y=x^3+\frac{3x^2}{2}+C_1x+C_2\\
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 4 — Special Second-Ordered Differential Equations

Find the general solution of the differential equation

y''+2y'-6x-3=0

Solution:

A second-order differential equation can be written in the form:

ay″ + by' + cy = g(x)

Therefore, the problem given is a second order linear equation.

Here are STEPS on how to get the general solution:

i. simplify the equation to a Second ODE Form

y''+2y'=6x+3

ii. Let 

\begin{align*}
y' = P=\frac{dy}{dx} 
\\
\\
and
\\
\\
y'' = \frac{dP}{dx}
\\

\\
\frac{dP}{dx}+P2\:=\:6x+3
\end{align*}

iii. By recalling, we can see that the equation is in First-order linear differential equation form. Solving the simplified equation using FOLDE.

\begin{align*}
P(x) & =2
\\
and
\\
Q(x) &=6x+3
\\
\\
\frac{dP}{dx}+P2\:& =\:6x+3
\end{align*}

Find the integrating factor

\begin{align*}
ɸ & =e^{\int \:P\left(x\right)dx}
\\
ɸ & =e^{\int \:\:2dx}
\\
ɸ & =e^{2x}
\end{align*}

Substituting the I.F. to the formula

\begin{align*}
Pɸ & =\int \:ɸQ\left(x\right)dx+C_1
\\
Pe^{2x} & =\int \:e^{2x}(6x+3)dx+C_1
\\
Pe^{2x} & =\int \:\left(e^{2x}6x+3e^{2x}\right)dx+C_1
\end{align*}

Integrating the first term

\begin{align*}
\int \:e^{2x}6xdx=

6\cdot \int \:xe^{2x}dx
\\
\end{align*}

Let u = 2x and du/2 = dx

\frac{3}{2}\int \:e^uudu

By IBP, Let v=u, dv=du and eudu, n=eu. 

\begin{align*}
nv-\int ndv &
\\
ue^u-\frac{3}{2}\int \:\:e^udu & = e^uu-e^u
\\
& =3e^{2x}x-\frac{3}{2}e^{2x}
\end{align*}

for the second term

\int \:3e^{2x}dx

Let u = 2x and du/2 = dx

\begin{align*}
\frac{3}{2}\int \:e^udu & =\frac{3}{2}e^u
\\
& =\frac{3}{2}e^{2x}
\end{align*}

Combining all the solved terms we get

\begin{align*}
Pe^{2x} & =3e^{2x}x-\frac{3}{2}e^{2x}+\frac{3}{2}e^{2x}+C_1
\\
Pe^{2x} & =3e^{2x}x+C_1
\end{align*}

Based on the equation that we derived it is now a separable differential equation, therefore,

\begin{align*}
&\left[\frac{dy}{dx}e^{2x}=3e^{2x}x+C_1\right]\frac{1}{e^{2x}}
\\
\frac{dy}{dx} & =3x+\frac{C_1}{e^{2x}}
\\
\int \:dy & =\int \:\left(3x+\frac{C_1}{e^{2x}}\right)dx+C_2
\end{align*}

GENERAL SOLUTION:

y=\frac{3x^2}{2}-\frac{C_1e^{-2x}}{2}+C_2
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Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 9 Problem 5 — Special Second-Ordered Differential Equations

Find the general solution of the differential equation

yy''+2\left(y\right)^2=0

Solution:

Based on Special Second-Ordered Differential Equation: Special case 3

F\left(\frac{d^2y}{dx^2},\:\frac{dy}{dx},\:y\right)=0

Denote and substitute to the given equation.

P= y' =\frac{dy}{dx}  \\ P\frac{dp}{dy}= y'' =\frac{d^2y}{dx^2}

We will have,

y(P\frac{dp}{dx})+2(P)^2=0

Divide both sides with

 \:\frac{1}{yP}

We will come to,

\frac{dp}{dy}+\frac{2P}{y}=0

Tranpose, 

\frac{2P}{y}

We will have

\frac{dp}{dy}=-\frac{2P}{y}

Integrate both sides,

\int \frac{dp}{dy}=-\int\frac{2P}{y}

The equation will become a SEPARABLE DIFFERENTIAL EQUATION, multiply both sides with

\frac{dy}{P}\:

We will come to the equation:

 \frac{dp}{P}=-\frac{2}{y}dy

Integrate both sides,

\int \frac{dp}{P}=-\int\frac{2}{y}dy

The answer will be:

\ln \left(P\right)=\ln \left(y^{-2}\right)+lnC

Apply logarithmic definition and exponent rule

loga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c

The answer will be:

P=\frac{C}{y^2}

Recall that

P=\frac{dy}{dx}

Substitute the original value of P,

\frac{dy}{dx}=\frac{C}{y^2}

Again, this is a Separable Differential Equation, multiply both sides with:

y^{2}dx

It will become

y^{2}dy=Cdx

Integrate both sides,

\int y^{2}dy=\int Cdx

The answer will be

\frac{y^3}{3}=C1x+C2

Multiply both sides with 3 and the final answer will be

y^3=C_1x+C_2

You can still solve it explicitly,

y=\sqrt[3]{C_1x+C_2}
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 6 — Special Second-Ordered Differential Equations

Solve the following differential equation

yy''-\left(y'\right)^2+y'=0

Solutions:

Basically, We need to make the orders of each term to 1. To be able to further break down the equation.

so\:we\:let\:P=y' \\
\:\:\:\:\:\:\:\:\:\:\:\:\:P\frac{dp}{dy}=y''

Substituting to the equation, we get 

yP\frac{dP}{dy}-P^2+P=0

Removing the variables y and P from the 1st term we get 

\frac{dP}{dx}-\frac{P}{y}+\frac{1}{y}=0\:\:\:\:\:\: , the \:equation\:has\:become\:a \:FOLDE\\
\;\\
\frac{dP}{dy}-\frac{P}{y}=-\frac{1}{y}\:\:\:\:\:\:\:\:\:\:\:T\left(y\right)=-\frac{1}{y},\:\:\:\:\:Q\left(y\right)=-\frac{1}{y}\\
\:\\\:\:\:\:
\phi =e^{\int \:-\frac{1}{y}dy}\\
=y^{-1}
\\\:\:\:\:\:\:\:\:\:P\phi=\int\phi\:Q(y)dy\:+\:C_{1}
\\\:\:\:\:\:\:\:\:\:\:\:\:Py^{-1}=\int \:y^{-1}\left(-y^{-1}\right)dy+C_{1}
\\
Py^{-1}=\int \:-y^{-2}dy+C_{1}
\\
\:\\
\frac{P}{y}=\frac{1}{y}+C_{1}
\\\:
\:\\\:\:
P=1+yC_{1}
\\
\frac{dy}{dx}=1+yC_{1}

By means of Separation of Variables

\\
\frac{dy}{1+yC_{1}}=\:dx
\\ 
\int \:\frac{dy}{1+yC_{1}}=\int \:dx
\\\:\:\:\:\:\:\:\:
let\:u=1+yC_{1}
\\\:\:\:\:\:\:
du=C_{1}dy\\
\frac{du}{C_{1}}=dy
\\\:\:\:
\frac{1}{C_{1}}\int \:\frac{du}{u}=\int \:dx
\\\:\:\:\:\:\:\:\:\:\:\:
\frac{1}{C_{1}}ln\:u=x+C_{2}

We get

ln\left|1+yC_{1}\right|=C_{1}x+C_{2}
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 2 — Special Second-Ordered Differential Equations

Question:

\frac{d}{dx}\left(x\:\frac{dy}{dx}\:+\left(1+x\right)\:y\right)\:=12\:\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1

Solution:

Since the equation is in the form d/dx (dy/dx + y P(x) = Q(x) , we use Case 1

\begin{align*}
let\:u & = x\frac{dy}{dx}+\left(1+x\right)\:y
\\ \int \:\frac{du}{dx}\:&=\int \:12
\\u &=12x+C_1
\end{align*}

Solving for the value of C1 using the initial values. 

\begin{align*}
x\frac{dy}{dx}+\left(1+x\right)y\: & =12x\:+C_1\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1
\\1(0) + (1+1)0 & = 12(1) + C_1  
\\0+0 & =12+C_1
\\-12 & =C_1
\end{align*}

Rewriting the equation into the general form of a first-order linear differential equation (FOLDE).

\left[x\frac{dy}{dx}+\left(1+x\right)y=12x-12\right]\frac{1}{x}
\\\frac{dy}{dx}+\left(\frac{1+x}{x}\right)y=12-\frac{12}{x}
\\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

Since the equation is now in the form of dy/dx + y P(x) = Q(x), we use FOLDE

\\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

From the general form of a first-order differential equation, we have

\\P \left( x \right)= \left(\frac{1}{x}+1\right) 
\\Q\left( x \right)= 12-\frac{12}{x}

Compute for the integrating factor

\begin{align*}
\phi &= e^ {\int P\left(x \right) dx}
\\\phi & =e^{\int \:\left(\frac{1}{x}+1\right)dx}
\\\phi & \:=e^{\ln x+x}
\\\phi & \:=x\left(e^x\right)
\end{align*}

Substituting everything to the solution of a first-order linear differential equation, we have

y(xe^x)=\int xe^x\left(12-\frac{12}{x}\right)dx+C_2
\\y\left(xe^x\right)=\int \left(12xe^x-12e^x\right)dx+C_2
\\yxe^x=\int 12xe^x-\int \:12e^x\:dx+C_2​

Use Integration by Parts to solve for the first integral

\begin{align*}
\int \:12xe^xdx & = 12 \int xe^x dx\\ 
 u = x &  &du=dx  \\
dv = & e^x \  & v=e^x  \\
\text{Therefore} \\ 
uv-\int \:vdu & =xe^x-\int \:e^xdx \\
& =xe^x-e^x \\
\text{Consequently} \\
\int \:12xe^xdx & = 12 \left( xe^x-e^x\right)
\end{align*}

Therefore,

\begin{align*}
yxe^x & =12\left(xe^x-e^x\right)-12e^x+C_2
\\yxe^x &=12xe^x-12e^x-12e^x+C_2
\\yxe^x &=12xe^x-24e^x+C_2
\end{align*}

Solving for C2

\begin{align*}
yx & =12x-24+\frac{C_2}{e^x}\:\:;\:y=0,\:x=1
\\0\left(1\right) & =12\left(1\right)-24+\frac{C_2}{e^1}\:
\\0 & =12-24e+\frac{C_2}{e^1}\:
\\0 &=-12+\frac{C_2}{e^1}\:
\\12e^1 & =C_2\:
\end{align*}

Therefore, the solution to the problem is

yx=12x-24+\frac{12e^1}{e^x}
\\or
\\yx=12x-24+12e^{1-x}

Determining if a given Differential Equation is Separable or Not

Determine whether each of the following differential equations is or is not separable, and, if it is separable, rewrite the equation in the form dy/dx=f(x) g(y).
\qquad \textbf{a}) \quad \frac{dy}{dx}=xy-3x-2y+6
\qquad \textbf{b})\quad \frac{dy}{dx}=\sin \left( x+y \right)
\qquad \textbf{c}) \quad y\frac{dy}{dx}=e^{x-3y^2}

Solution:

Part A

\begin{align*}
\frac{dy}{dx} & = xy-3x-2y+6 \\
\frac{dy}{dx} & = \left( xy-3x \right)-\left( 2y-6 \right)\\
\frac{dy}{dx} & = x\left( y-3 \right)-2\left( y-3 \right)\\
\frac{dy}{dx} & = \left( x-2 \right)\left( y-3 \right)
\end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.

Part B

\begin{align*}
\frac{dy}{dx} & = \sin\left( x+y \right) \\
\frac{dy}{dx} & = \sin\left( x \right)\cos\left( y \right) +\cos\left( x \right)\sin\left( y \right)\\
\end{align*}

Since F(x,y) is not factorable in the form f(x) g(y), the given differential equation is not separable.

Part C

\begin{align*}
y \frac{dy}{dx} & = e^{x-3y^2}\\
y \frac{dy}{dx} & = \frac{e^x}{e^{3y^2}}\\
\frac{dy}{dx} & =\frac{e^x}{y e^{3y^2}} \\
\frac{dy}{dx} & = e^x \left( \frac{1}{ye^{3y^2}} \right) \\
\end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.