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

In summary, the solution to the differential equation dB/dt+6B=40 with B(1)=90 is B(t)=\frac{10\left(2+25e^{6(1-t)} \right)}{3}.
  • #1
MarkFL
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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.
 
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  • #2
Hello Clare,

We are given the IVP:

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

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:

\(\displaystyle \frac{dB}{dt}=40-6B\)

\(\displaystyle \frac{3}{3B-20}\,dB=-6dt\)

Integrate:

\(\displaystyle \int \frac{3}{3B-20}\,dB=-6\int\,dt\)

\(\displaystyle \ln|3B-20|=-6t+C\)

Convert from logarithmic to exponential form:

\(\displaystyle 3B-20=Ce^{-6t}\)

Hence:

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

ii) Linear equation - integration factor:

\(\displaystyle \frac{dB}{dt}+6B=40\)

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

\(\displaystyle e^{6t}\frac{dB}{dt}+6e^{6t}B=40e^{6t}\)

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

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

Integrate with respect to $t$:

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

\(\displaystyle e^{6t}B=\frac{20}{3}e^{6t}+C\)

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

Rewriting the parameter $C$, we obtain:

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

iii) Linear inhomogeneous equation:

\(\displaystyle \frac{dB}{dt}+6B=40\)

Observing the characteristic root is:

\(\displaystyle r=-6\)

we find then that the homogeneous solution is:

\(\displaystyle B_h(t)=c_1e^{-6t}\)

We then look for a particular solution of the form:

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

Substitution into the ODE gives us:

\(\displaystyle 6A=40\implies A=\frac{20}{3}\)

And so by superposition, we find:

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

Rewriting the parameter $c_1$, we obtain:

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

Finding the value of the parameter:

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

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

\(\displaystyle 20+Ce^{-6}=270\)

\(\displaystyle Ce^{-6}=250\)

\(\displaystyle C=250e^{6}\)

The Solution:

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

\(\displaystyle B(t)=\frac{20+250e^{6}e^{-6t}}{3}=\frac{10\left(2+25e^{6(1-t)} \right)}{3}\)
 

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

1. What is a first order linear IVP?

A first order linear IVP (initial value problem) is a type of differential equation that involves a first derivative of a function, along with an initial condition that specifies the value of the function at a certain point. The general form of a first order linear IVP is y' = f(x) + g(x)y, where f(x) and g(x) are functions of x.

2. How do you solve a first order linear IVP?

To solve a first order linear IVP, you can use the method of separation of variables or the integrating factor method. Both methods involve manipulating the equation to isolate the dependent variable and then integrating both sides to find the general solution. The initial condition can then be used to find the specific solution for the given problem.

3. What is the purpose of solving a first order linear IVP?

Solving a first order linear IVP allows us to find the general solution to a differential equation and then use the initial condition to find the specific solution for a given problem. This can be useful in many fields of science, such as physics and engineering, where differential equations are commonly used to model real-world situations.

4. Can all first order differential equations be solved using the methods for first order linear IVPs?

No, not all first order differential equations can be solved using the methods for first order linear IVPs. Other types of first order differential equations include separable, exact, and homogeneous equations, which require different methods of solving.

5. What are some applications of first order linear IVPs?

First order linear IVPs have many applications in various fields of science, including physics, engineering, and biology. They can be used to model processes such as population growth, chemical reactions, and electrical circuits. They are also useful in solving problems involving rates of change and finding optimal solutions.

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