MHB Clare's question at Yahoo Answers regarding a first order linear IVP

AI Thread Summary
The discussion focuses on solving the initial value problem (IVP) for the differential equation dB/dt + 6B = 40 with the initial condition B(1) = 90. Various methods are presented, including separation of variables, using an integrating factor, and identifying a particular solution for the linear inhomogeneous equation. Each method leads to the general solution B(t) = (20 + Ce^(-6t))/3, where C is determined using the initial condition. Ultimately, the specific solution is found to be B(t) = (10(2 + 25e^(6(1-t))))/3.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Find the solution to the differential equation dB/dt+6B=40?

Find the solution to the differential equation
dB/dt+6B=40,
with B(1)=90.

I have posted a link there to this thread so the OP can view my work.
 
Mathematics news on Phys.org
Hello Clare,

We are given the IVP:

$$\frac{dB}{dt}+6B=40$$ where $$B(1)=90$$

There are several ways we could go about solving the ODE associated with the given IVP.

i) Separable equation:

We may write the ODE as:

$$\frac{dB}{dt}=40-6B$$

$$\frac{3}{3B-20}\,dB=-6dt$$

Integrate:

$$\int \frac{3}{3B-20}\,dB=-6\int\,dt$$

$$\ln|3B-20|=-6t+C$$

Convert from logarithmic to exponential form:

$$3B-20=Ce^{-6t}$$

Hence:

$$B(t)=\frac{20+Ce^{-6t}}{3}$$

ii) Linear equation - integration factor:

$$\frac{dB}{dt}+6B=40$$

Multiply through by the integrating factor $\mu(t)=e^{6t}$:

$$e^{6t}\frac{dB}{dt}+6e^{6t}B=40e^{6t}$$

Observing the left side is the differentiation of a product, we obtain:

$$\frac{d}{dt}\left(e^{6t}B \right)=40e^{6t}$$

Integrate with respect to $t$:

$$\int\,d\left(e^{6t}B \right)=\frac{20}{3}\int e^{6t}\,6\,dt$$

$$e^{6t}B=\frac{20}{3}e^{6t}+C$$

$$B=\frac{20}{3}+Ce^{-6t}$$

Rewriting the parameter $C$, we obtain:

$$B(t)=\frac{20+Ce^{-6t}}{3}$$

iii) Linear inhomogeneous equation:

$$\frac{dB}{dt}+6B=40$$

Observing the characteristic root is:

$$r=-6$$

we find then that the homogeneous solution is:

$$B_h(t)=c_1e^{-6t}$$

We then look for a particular solution of the form:

$$B_p(t)=A\implies \frac{dB}{dt}=0$$

Substitution into the ODE gives us:

$$6A=40\implies A=\frac{20}{3}$$

And so by superposition, we find:

$$B(t)=B_h(t)+B_p(t)=c_1e^{-6t}+\frac{20}{3}$$

Rewriting the parameter $c_1$, we obtain:

$$B(t)=\frac{20+Ce^{-6t}}{3}$$

Finding the value of the parameter:

Now, using the initial value, we may determine that value of the parameter:

$$B(1)=\frac{20+Ce^{-6}}{3}=90$$

$$20+Ce^{-6}=270$$

$$Ce^{-6}=250$$

$$C=250e^{6}$$

The Solution:

And so, we find the solution to the IVP is then:

$$B(t)=\frac{20+250e^{6}e^{-6t}}{3}=\frac{10\left(2+25e^{6(1-t)} \right)}{3}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top