Solution to Initial Value Problem: dy/dx = y-3, y(0)=4

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Discussion Overview

The discussion revolves around solving the initial value problem defined by the differential equation dy/dx = y - 3 with the initial condition y(0) = 4. The focus is on the separation of variables and the integration process involved in finding the solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants express difficulty in separating the variables in the equation dy/dx = y - 3 and question whether it is separable.
  • One participant suggests starting with the separation of variables and provides the integration steps, indicating that ln(y - 3) = x + C is the result of integrating both sides.
  • Another participant acknowledges understanding the integration process and notes the relationship between the natural logarithm and the exponential function, leading to the expression y = Ce^x + 3.
  • There is a suggestion to apply the initial condition to determine the constant C, although the specific value is not calculated in the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the steps to solve the problem, but there is no consensus on the final solution since the constant C has not been explicitly determined in the discussion.

Contextual Notes

The discussion does not resolve the specific value of the constant C, which depends on applying the initial condition y(0) = 4. The steps leading to the solution are presented, but the final expression remains incomplete without this application.

Mitchtwitchita
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The solution of the initial value problem, dy/dx = y-3. y(0) = 4, what does y=?

dy/y-3 = dx
1/y-3 dy = dx
ln(y-3) = ? + C

I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.
 
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Mitchtwitchita said:
The solution of the initial value problem, dy/dx = y-3. y(0) = 4, what does y=?

dy/y-3 = dx
1/y-3 dy = dx
ln(y-3) = ? + C

I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.

well you started it right:

\frac{dy}{dx}=y-3=>\frac{dy}{y-3}=dx

Now to get our answer we integrate both parts

\int\frac{dy}{y-3}=\int dx=>ln(y-3)=x+C

Now do you know how to isolate y by itself. Hint: remember that ln and the exponential function with base "e" are inverses. So, exponentiate both sides, and try to go as further as you can. If you're still stuck, ask again.
 
I do. thanks stupidmath. I guess it was the "invisible" x that got me.

y-3 = Ce^x
y = Ce^x + 3
 
Mitchtwitchita said:
I do. thanks stupidmath. I guess it was the "invisible" x that got me.

y-3 = Ce^x
y = Ce^x + 3

Now apply your initial conditons to find C.
 

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