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Calculus 2 Differential Equation?
a)Find the solution of the differential equation dz/dt + e^(t+z)=0 that satisfies the condition z(8)=-8
Can someone please help I really dont understand :/ can you please include the steps too i will choose best answer so you can get 10 pts
Also:
b) find the solution of the differential equation xy'+y=y^2 that satisfies the initial condition y(4)=-1
thanks!
1 Answer
- ?Lv 77 years ago
a) e^(t + z) = e^(t)*e^(z)
dz/dt = -e^(t)*e^(z)
separate variables
e^(-z)dz = -e^(t) integrate each side
-e^(-z) = -e^(t) + k
z = -t + C
When t = 8, z = -8
-8 = -8 + C, so C = 0
z = -t
Check: dz/dt = -1, e^(t + z) = e^0 = +1
b) xy' + y = y^2,
x*dy/dx = y^2 - y
separate variables
[1/(y^2 - y)]dy = [1/(y - 1) - 1/y)]dy = (1/x)dx
ln(y - 1) - ln(y) = ln[(y - 1)/y] = ln(x) + C ......(1)
Initial condition equivalent to y = -1 when x = 4
Avoid having to use ln of a negative by writing
ln[(1 - 1/y] = ln(x) + C ..................................(2)
ln[(1 - 1/(-1)] = ln(4) + C
C = ln 2 - 2ln2 = - ln 2
ln[(y - 1)/y] = ln(x) - ln 2 = ln(x/2)
(y - 1)/y = x/2
y = -2/(x - 2)
Regards - Ian H