Solve the separable differential equation

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Homework Help Overview

The discussion revolves around solving a separable differential equation given by dx/dy = -0.6y with an initial condition y(0) = 5. Participants are examining the steps taken to integrate and solve for the function y.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the equation through integration but expresses uncertainty about the correctness of their solution. Some participants question whether the original differential equation was stated correctly, suggesting a possible typo regarding the variables involved.

Discussion Status

Participants are actively engaging with the original poster's approach, offering clarifications and corrections regarding the integration process. There is a recognition of the need to solve for the function before applying initial conditions, and some guidance has been provided to help clarify the misunderstanding.

Contextual Notes

There is a noted ambiguity regarding the formulation of the differential equation, with participants questioning whether dx/dy is indeed correct or if it should be dy/dx. This uncertainty may affect the interpretation of the problem.

hardatwork
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Homework Statement



dx/dy=-0.6y
y(0)=5

Homework Equations





The Attempt at a Solution


I tried solving it by
\intdy/y=\int-0.6dx
ln(y)=-0.6x+c
ln(y(0))=-0.6(0)+c
ln(5)=c
ln(y)=-0.6x+ln(5)
y=e^{-0.6x}+5
But its incorrect. I don't know what I am doing wrong. Can someone helping see what I am doing wrong? Thank You so much!
 
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You should solve for your function before substituting initial conditions. Remember that e^{ln|y|}=y
 
hardatwork said:

Homework Statement



dx/dy=-0.6y

Is this a typo? Do really mean dx/dy=-0.6y, or do you mean dy/dx=-0.6y?

\intdy/y=\int-0.6dx
ln(y)=-0.6x+c
ln(y(0))=-0.6(0)+c
ln(5)=c
ln(y)=-0.6x+ln(5)
y=e^{-0.6x}+5
But its incorrect. I don't know what I am doing wrong. Can someone helping see what I am doing wrong? Thank You so much!

If \ln y=-0.6x+\ln 5, then

y=e^{-0.6x+\ln 5}=e^{\ln 5}e^{-0.6x}=5e^{-0.6x}\neq e^{-0.6x}+5
 
Okay. That makes so much sense. Thank You so much
 

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