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

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
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:

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