Question

Solve the initial-value problem.

dr/dt + 2tr = r, r(0) = 5

Answer 1

$\displaystyle\frac{dr}{dt} + 2tr = r\ \Rightarrow\ \frac{dr}{dt} = r - 2tr\ \Rightarrow\ \frac{dr}{dt} = r(1 - 2t) \ \Rightarrow \\ \\ \frac{dr}{r} = (1 - 2t)\, dt\ \Rightarrow \int \frac{dr}{r} = \int(1 - 2t) \, dt\ \Rightarrow\ \ln|r| = t - t^2 + C. \\ \\ |r| = e^{t - t^2 + C} = ke^{t - t^2}.\ \text{Since } r(0) = 5, 5 = ke^0 = k. \\ \\ \text{Thus, } r(t) = 5e^{t - t^2}.$

Answer 2

The solution to given initial value problem $\frac{dr}{dt}+2tr=r$ , r(0) = 5 is $r=5.e^{t-t^2}$

What is initial value problem?

"It is a differential equation with some initial conditions."

What is differential equation?

"An equation with one or more derivatives of a function"

For given question,

We have been given a differential equation dr/dt + 2tr = r, and initial condition r(0) = 5

We need to solve given initial-value problem.

Consider given differential equation,

$\Rightarrow \frac{dr}{dt}+2tr=r\\\\\Rightarrow \frac{dr}{dt}=r-2tr\\\\\Rightarrow \frac{dr}{dt}=r(1-2t)\\\\\Rightarrow \frac{dr}{r}=(1-2t)dt\\\\\Rightarrow \int \frac{dr}{r} \, = \int (1-2t) dt \\\\\Rightarrow ln~r=t-t^2+c\\\\\Rightarrow r=e^{t-t^2+c}\\\\\Rightarrow r=e^c.e^{t-t^2}\\\\\Rightarrow r=m.e^{t-t^2}$

So, the general solution to given differential equation dr/dt + 2tr = r is,

$r=m.e^{t-t^2}$

From the initial condition r(0) = 5,

for t = 0, r = 5

$\Rightarrow 5=m.e^{0-0^2}\\\\\Rightarrow 5=m.1\\\\\Rightarrow m=5$

This means, the particular solution to given differential equation is  $r=5.e^{t-t^2}$

Therefore, the solution to given initial value problem $\frac{dr}{dt}+2tr=r$ , r(0) = 5 is $r=5.e^{t-t^2}$

Learn more about the initial value problem here:

https://brainly.com/question/8736446

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