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Can anyone solve this differential equations problem?
x'+2xe^(-4t)=0 Initial condition x(0)=5
I need the equation x(t)
2 Answers
- schmisoLv 71 decade agoFavorite Answer
The is a separable differential equation. So separate variables and integrate:
dx/dt + 2∙x∙e^(-4∙t) = 0
<=>
dx/dt = - 2∙x∙e^(-4∙t)
<=>
(1/x) dx = -2∙e^(-4∙t) dt
=>
∫ (1/x) dx = ∫ -2∙e^(-4∙t) dt
=>
ln(x) = (1/2)∙e^(-4∙t) + c
<=>
x = e^{ (1/2)∙e^(-4∙t) + c }
<=>
x = e^(c) ∙ e^{ (1/2)∙e^(-4∙t) }
set C = e^c
<=>
x = C∙e^{ (1/2)∙e^(-4∙t) }
Apply initial condition to find constant C:
x(0) = 5
<=>
C∙e^{ (1/2)∙e^(-4∙0) } = 5
<=>
C∙e^(1/2) = 5
<=>
C = 5/e^(1/2) = 5∙e^(-(1/2))
Hence:
x(t) = 5∙e^(-(1/2))∙∙e^{ (1/2)∙e^(-4∙t) }
= 5∙e^{ (1/2)∙e^(-4∙t) - (1/2) }
= 5∙e^{ (1/2)∙(e^(-4∙t) - 1) }
- eagleLv 51 decade ago
Yes, there must be at least one person who can solve your differential equations problem.
