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Equazioni differenziali
- y' + 3y - 1 = 0
y' + 3y = 1
y = e- 3 dx · [ (e3 dx · (1)) dx + c]
y = e-3x[ e3x dx + c]
y = e-3x[1/3 e3x + c]
y = 1/3 + c/e3x
www.matematicamente.it
y' + 3y = 1
y = e- 3 dx · [ (e3 dx · (1)) dx + c]
y = e-3x[ e3x dx + c]
y = e-3x[1/3 e3x + c]
y = 1/3 + c/e3x