I\'m trying to solve these two initial value problems and I have them mostly fig
ID: 2975145 • Letter: I
Question
I'm trying to solve these two initial value problems and I have them mostly figured out but I can't finish them...can anyone help?
here's what I have:
1. The population of field mice satisfies the differential equation dp/dt = .5p - 450, where t is measured in months. Find the time at which the population becomes extinct if p(0) = 850.
I separate the equation and after integrating both sides I get: .5ln|p-450| + C = t + C
2. Solve the initial value problem y' = 2y^2 + xy^2, y(0) = 1,
I separate the equation, integrate both sides and get -1/y + C = 2x + x^2/2 + C
where do I go from here? Thanks
Explanation / Answer
1 : 1st ; you integrated incorrectly ...2nd ; you only need one [ + C ] when integrating ;
{ p = 900 - 50 e^(t/2) }....{ ˜ 5.8 ]
#2. delete one [ C ] , then let x = 0 and y = 1 to find [ C ]
{ y = 1 / [ -2x - x² / 2 + 1 ] }
here its solved below
1. dp/dt = .5(p - 900)
dp / (p-900) = .5dt
ln|p-900| = .5t + c
p - 900 = C(e^(.5t))
p = 900 + Ce^(.5t)
p(0) = 850 = 900 + C
C = -50
p = 900 - 50e^(.5t)
0 = 900 - 50e^(.5t)
18 = e^(.5t)
ln18 = .5t
2ln18 = t
2. y' = y^2(2+x)
y'/y^2 = (2+x)
? dy / y^2 = ? 2+x dx
-1/y = 2x + (1/2)x^2 + C
-1/y = (4x + x^2 + C)/2
y = -(2) / (4x+x^2+c)
y(0) = 1 = -2/c
c = -2
y = -2 / (4x+x^2 - 2)