dB/dt = - 0.05B
dB/B = - 0.05dt
∫ dB/B =- ∫0.05dt
ln |B| = - 0.05t +k
|B| = e( -0.05t +k)
|B| = ek e-0.05
|B| = λ e-0.05t
B = ± λ e-0.05t
B = c e-0.05t
Afreen K.
asked 07/23/21Which function satisfies the equation below? (c is an arbitrary constant.)
dB/dt= (-0.05)B
dB/dt = - 0.05B
dB/B = - 0.05dt
∫ dB/B =- ∫0.05dt
ln |B| = - 0.05t +k
|B| = e( -0.05t +k)
|B| = ek e-0.05
|B| = λ e-0.05t
B = ± λ e-0.05t
B = c e-0.05t
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