Question

Find the solution to the differential equation dB/dt+4B=20 with B(1)=30

Answer 1

Answer:

The solution of the differential equation is $B=5+25e^{-4t+4}$

Step-by-step explanation:

The differential equation $\frac{dB}{dt}+4B=20$ is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

$y'(t)=g(t)\cdot h(y)$

where g(t) and h(y) are given functions.

We can rewrite our differential equation in the form of a first-order separable ODE in this way:

$\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4$

Integrating both sides

$\frac{1}{5-B}\frac{dB}{dt}=4\\\frac{1}{5-B}\cdot dB=4\cdot dt\\\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {4} \, dt$

The integral of left-side is:

$\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C$

The integral of right-side is:

$\int\limits {4} \, dt = 4t + C$

We can join the constants, and this is the implicit general solution

$-\ln \left|5-B\right|+C=4t + C\\-\ln \left|5-B\right|=4t + D$

If we want to find the explicit general solution of the differential equation

We isolate B

$-\ln \left|5-B\right|=4t + D\\\ln \left|5-B\right|=-4t+D\\\left|5-B\right|=e^{-4t+D}$

Recall the definition of |x|

$|x|=\left \{ {{x, \:if \>x\geq \>0 } \atop {-x, \:if \>x0}} \right.$

So

$\left|5-B\right|=e^{-4t+D}\\5-B= \pm \:e^{-4t+D}\\B=5 \pm \:e^{-4t+D}\\B=5\pm \:e^{-4t}\cdot e^{D}\\B=5+Ae^{-4t}$

where $A=\pm e^{D}$

Now B(1) =30 implies

$B=5+Ae^{-4t}\\30=5+Ae^{-4}\\30-5=Ae^{-4}\\25e^{4}=A$

And the solution is

$B=5+(25e^{4})e^{-4t}\\B=5+25e^{-4t+4}$