- Thread starter
- Admin
- #1
Clare's question at Yahoo! Answers regarding a first order linear IVP
- Thread starter MarkFL
- Start date
- Thread starter
- Admin
- #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:
\(\displaystyle B(t)=\frac{20+250e^{6}e^{-6t}}{3}=\frac{10\left(2+25e^{6(1-t)} \right)}{3}\)
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:
\(\displaystyle B(t)=\frac{20+250e^{6}e^{-6t}}{3}=\frac{10\left(2+25e^{6(1-t)} \right)}{3}\)