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1 answer:

3 0

Yes, it does. A good way to fill another 6 characters is with a "please".

Watch carefully:

You have said that 2x - 5 = 7

Add 5 to each side: 2x = 12

Divide each side by 2: x = 6 .

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Step-by-step explanation:

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3 years ago

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

natita [175]

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:

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