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