Evaluate the initial-value problem. dr/dt + 2tr = r. r(0)=5?

Question

what would the answer be if C were calculated for? i get r = e^(t-t^2+ln5).

so that equals 5e^(t-t^2). Thanks!

Answer 1

Use separation of variables:
dr/dt + 2tr = r
Subtract 2tr from both sides:
dr/dt = r - 2tr
Take out an r on the right hand side:
dr/dt = r(1 -2t)
Cross-multiply:
dr / r = (1-2t)dt

Now integrate both sides:
∫dr / r = ∫(1-2t)dt
ln(r) + C = t - t^2 + D

Combine the arbitrary constants:
ln(r) = t - t^2 + C

We want to solve for r, so let both sides be exponents with base e:
e^(ln(r)) = e^(t - t^2 + C)
r = e^(t - t^2 + C)

Now plug in the initial value to find the value of C:
r(0) = 5
r(0) = 5 = e^(5 - 5^2 + C)
Take the ln of both sides:
ln(5) = ln[e^(-20 + C)]
ln(5) = -20 + C
C = ln(5) + 20

So the final answer is:
r = e^(t - t^2 + ln(5) + 20)

Check the answer:
dr/dt = [e^(t - t^2 + ln(5) + 20)] * (1 - 2t)
And
dr/dt + 2tr = r
[e^(t - t^2 + ln(5) + 20)] * (1 - 2t) + 2t*(e^(t - t^2 + ln(5) + 20)) = e^(t - t^2 + ln(5) + 20)
Divide both sides by e^(t - t^2 + ln(5) + 20):
(1 - 2t) + 2t = 1
1 = 1
So the answer checks.



edit:
Scarlet, using an integrating factor with a nonlinear ODE is a bad idea, though your solution does work in this case.

Answer 2

dr/dt + 2tr = r
<=> dr/dt + (2t-1)r = 0
Integrating factor is e^∫(2t-1)dt = e^(t^2-t)
So we get
e^(t^2-t) dr/dt + (2t-1) e^(t^2-t) r = 0
<=> d/dt (e^(t^2-t) r) = 0
<=> e^(t^2-t) r = c
<=> r = c e^(t-t^2)
r(0) = c e^0 = c = 5, so we get
r(t) = 5 e^(t-t^2).

Answer 3

r(t)=5*exp( t - (t^2)/2 )