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  • 3.9: Differential Equations
    https://math.libretexts.org/Courses/Butler_Community_College/MA148%3A_Calculus_with_Applications_-_Butler_CC/03%3A_The_Integral/3.09%3A_Differential_Equations
    Start by multiplying by 0.02. \[ \begin{align*} \ln|0.02B-20| & = 0.02t+0.02C &\\ \ln|0.02B-20| & = 0.02t+D &\qquad\text{We can rename D=0.02C for simplicity.}\\ e^{\ln|0.02B-20|} & = e^{0.02t+D...Start by multiplying by 0.02. \[ \begin{align*} \ln|0.02B-20| & = 0.02t+0.02C &\\ \ln|0.02B-20| & = 0.02t+D &\qquad\text{We can rename D=0.02C for simplicity.}\\ e^{\ln|0.02B-20|} & = e^{0.02t+D} &\qquad\text{Exponentiate both sides: eleft=eright.}\\ |0.02B-20| & = e^{0.02t+D} &\qquad\text{Use the log rule eln(A)=A.}\\ 0.02B-20 & = e^{0.02t+D} &\qquad\text{Since the RHS is always positive, we can drop the abs value.}\\ 0.02B-20 & = e^{0.02t}e^D &\qquad\…
  • 3.8: Differential Equations
    https://math.libretexts.org/Courses/Penn_State_University_Greater_Allegheny/MATH_110%3A_Techniques_of_Calculus_I_(Gaydos)/03%3A_The_Integral/3.08%3A_Differential_Equations
    Start by multiplying by 0.02. \[ \begin{align*} \ln|0.02B-20| & = 0.02t+0.02C &\\ \ln|0.02B-20| & = 0.02t+D &\qquad\text{We can rename D=0.02C for simplicity.}\\ e^{\ln|0.02B-20|} & = e^{0.02t+D...Start by multiplying by 0.02. \[ \begin{align*} \ln|0.02B-20| & = 0.02t+0.02C &\\ \ln|0.02B-20| & = 0.02t+D &\qquad\text{We can rename D=0.02C for simplicity.}\\ e^{\ln|0.02B-20|} & = e^{0.02t+D} &\qquad\text{Exponentiate both sides: eleft=eright.}\\ |0.02B-20| & = e^{0.02t+D} &\qquad\text{Use the log rule eln(A)=A.}\\ 0.02B-20 & = e^{0.02t+D} &\qquad\text{Since the RHS is always positive, we can drop the abs value.}\\ 0.02B-20 & = e^{0.02t}e^D &\qquad\…
  • 5.9: Differential Equations
    https://math.libretexts.org/Courses/Chabot_College/MTH_15%3A_Applied_Calculus_I/05%3A_The_Integral/5.09%3A_Differential_Equations
    Start by multiplying by 0.02. \[ \begin{align*} \ln|0.02B-20| & = 0.02t+0.02C &\\ \ln|0.02B-20| & = 0.02t+D &\qquad\text{We can rename D=0.02C for simplicity.}\\ e^{\ln|0.02B-20|} & = e^{0.02t+D...Start by multiplying by 0.02. \[ \begin{align*} \ln|0.02B-20| & = 0.02t+0.02C &\\ \ln|0.02B-20| & = 0.02t+D &\qquad\text{We can rename D=0.02C for simplicity.}\\ e^{\ln|0.02B-20|} & = e^{0.02t+D} &\qquad\text{Exponentiate both sides: eleft=eright.}\\ |0.02B-20| & = e^{0.02t+D} &\qquad\text{Use the log rule eln(A)=A.}\\ 0.02B-20 & = e^{0.02t+D} &\qquad\text{Since the RHS is always positive, we can drop the abs value.}\\ 0.02B-20 & = e^{0.02t}e^D &\qquad\…

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