- Thread starter
- Admin
- #1

I have posted a link there to this thread so the OP can view my work.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.

- Thread starter MarkFL
- Start date

- Thread starter
- Admin
- #1

I have posted a link there to this thread so the OP can view my work.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.

- Thread starter
- Admin
- #2

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.

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:

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

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

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:

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

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

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:

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

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

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

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

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

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

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

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