Author Archives: Via-Kathrina Cadotdot

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

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

Using,

Vbrine+(rate of brine out)tdSdt=8LM(S400+4t)=8S(400+4t)dSdt=2S(100+t)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:

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

Solve for the integrating factor using the formula:

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

Apply,

σ=e∫ 2100+tdtσ=e2ln(100+t)σ =eln(100+t)2σ =(100+t)2\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:

Sσ=∫ σQ(t)dt+CS\sigma =\int \:\sigma Q\left(t\right)dt+C

Apply,

S(100+t)2=∫ (100+t)230dt+CS(100+t)2=30∫ (100+t)2dt+CS(100+t)2=30 (100+t)33dt+CS(100+t)2=10(100+t)3+C→eqn.1S\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(100+60) 2 =10(100+60) 3C=2NL ; C=S(400+4t)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=S(400+4t)\:C=\frac{S}{\left(400+4t\right)}

Isolate S,

S=C(400+4t); C=2, t=60S=2(400+4(60))S=1280NS=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(100+t)2=10(100+t)3+C;S=1280,t=601280(100+60)2=10(100+60)3+C32768000=40960000+C32768000−40960000=CC=−8192000S\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(100+t)2=10(100+t)3−8192000S\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(100+t)2=10(100+t)3−8192000;t=0S(100+0)2=10(100+0)3−8192000S(1000)21000=18080001000S=180.8NS\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 9 Problem 5 — Special Second-Ordered Differential Equations

Find the general solution of the differential equation

yy′′+2(y)2=0yy''+2\left(y\right)^2=0

Solution:

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

F(d2ydx2, dydx, y)=0F\left(\frac{d^2y}{dx^2},\:\frac{dy}{dx},\:y\right)=0

Denote and substitute to the given equation.

P=y′=dydxPdpdy=y′′=d2ydx2P= y' =\frac{dy}{dx} \\ P\frac{dp}{dy}= y'' =\frac{d^2y}{dx^2}

We will have,

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

Divide both sides with

 1yP \:\frac{1}{yP}

We will come to,

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

Tranpose,

2Py\frac{2P}{y}

We will have

dpdy=−2Py\frac{dp}{dy}=-\frac{2P}{y}

Integrate both sides,

∫dpdy=−∫2Py\int \frac{dp}{dy}=-\int\frac{2P}{y}

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

dyP \frac{dy}{P}\:

We will come to the equation:

dpP=−2ydy \frac{dp}{P}=-\frac{2}{y}dy

Integrate both sides,

∫dpP=−∫2ydy\int \frac{dp}{P}=-\int\frac{2}{y}dy

The answer will be:

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

Apply logarithmic definition and exponent rule

logab=c then, b=acab+c=abacloga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c

The answer will be:

P=Cy2P=\frac{C}{y^2}

Recall that

P=dydxP=\frac{dy}{dx}

Substitute the original value of P,

dydx=Cy2\frac{dy}{dx}=\frac{C}{y^2}

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

y2dxy^{2}dx

It will become

y2dy=Cdxy^{2}dy=Cdx

Integrate both sides,

∫y2dy=∫Cdx\int y^{2}dy=\int Cdx

The answer will be

y33=C1x+C2\frac{y^3}{3}=C1x+C2

Multiply both sides with 3 and the final answer will be

y3=C1x+C2y^3=C_1x+C_2

You can still solve it explicitly,

y=C1x+C23y=\sqrt[3]{C_1x+C_2}
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