Question

Assume a is a non-zero constant, and use C for any constant of integration that you may have in your answer. dp/dt=2p+a solve differential equation p=

Answer 1

Answer:

$p=\dfrac{1}{2}(e^{2t + C'}-a)$

Step-by-step explanation:

We are given the following differential equation:

$\displaystyle\frac{dp}{dt} = 2p + a$

where a is a non zero constant.

Solving the differential equation:

$\displaystyle\frac{dp}{dt} = 2p + a\\\\\frac{dp}{2p+a} = dt\\\\\text{Integrating both sides}\\\\\int \frac{dp}{2p + a} = \int dt\\\\\frac{1}{2}\ln(2p+a) = t + C\\\\\text{where C is the constant of integration.}\\\\\ln(2p+a) = 2(t+C)\\\\2p+a = e^{2t + C'}\\2p = e^{2t + C'}-a\\\\p=\dfrac{1}{2}(e^{2t + C'}-a)$

Answer 2

After the solving the differential equation we get the value of p is

$P=\frac{1}{2} (e^{2t+C}-a)$

We are given the following differential equation:

$\frac{dp}{dt}=2p+a$

where a is a non-zero constant.

What is the differential equation?

A differential equation is an equation that relates one or more unknown functions and their derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two.

Solving the differential equation:

$\frac{dp}{dt}=2p+a$

$\frac{dp}{2p+a}=dt$

integrating both sides we get,

$\int{\frac{dp}{2p+a}=\int dt$

$\frac{1}{2} ln(2p+a)=t+C$

$2p+a=e^{2t+C}\\2p=e^{2t+C}-a\\p=\frac{1}{2} (e^{2t+C}-a)$

Therefore After the solving the differential equation ,we get the value of p is,

$P=\frac{1}{2} (e^{2t+C}-a)$

To learn more about the differential equation visit:

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