$Differential Equations- Initial Value Problem$

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$Differential Equations- Initial Value Problem$

I have my general solution but the initial condition is $y(0)=y_0$. How do I know what $y_0$ is?
\[y'=-y+5, y(0)=y_0. \]

Ordinary Differential Equations

Horta Emi

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The constant $y_0$ is an initial condition that should be given. Indeed $y_0$ can be any real number.

To find a general solution of $y'=-y+5$, we proceed as follows
\[\frac{dy}{dx}=-y+5 \Rightarrow \frac{dy}{-y+5}=dx.\]
Integrating both sides we get
\[- \ln (-y+5)=x+c. (*) \]
Substituting $x=0$ we get
\[- \ln (-y(0)+5)=0+c \Rightarrow c=-\ln(-y_0+5). \]
Substituting $c$ in (*)
\[- \ln (-y+5)=x-\ln(-y_0+5). \]
\[\Rightarrow \ln (\frac{1}{-y+5})=x+\ln(\frac{1}{-y_0+5}). \]
\[\Rightarrow \exp( \ln (\frac{1}{-y+5}) ) = \exp( x+\ln(\frac{1}{-y_0+5})) \]
\[\Rightarrow \frac{1}{-y+5}= \frac{1}{-y_0+5}\exp( x) \]
\[\Rightarrow -y+5=(-y_0+5)e^{-x}\]
\[\Rightarrow y=5-(-y_0+5)e^{-x} (**)\]
Here $y_0=y(0)$ can be any constant that should be given.

Equation (**) gives the solution of the ODE. If you notice, the problem says "plot the solutions for several values of y_0". Indeed $y_0$ can be taken to be any constant. For example if you take $y_0=1$, then the solution is
\[y=5-(-1+5)e^{-x}=5-4e^{-x}. \]
If you let $y_0=3$, then you will get
\[y=5-(-3+5)e^{-x}=5-2e^{-x}. \]
As I mentioned the initial condition $y_0$ should be chosen and can not be computed. Here we can freely choose $y_0$ to be any real number and get the corresponding solution.

Savionf

Savionf

0

Let me know if you need any clarifications.
- Horta Emi
  
  0
  
  I have my general solution then, but my constant isn’t given. I’m just given that x=0 but I need to solve for c but I can’t do that because y0 isn’t a number?
Savionf

0

y_0 has to be given and it can not be solved. Do you have any other information given?
- Horta Emi
  
  0
  
  The question quotes “Solve each of the following initial value problems and plot the solutions for several values of Yo. Then describe in a few words how the solutions resemble, and differ from, each other.” I would assume that I’m assuming different values for y(0) and drawing a direction field.
Savionf

0

y_0 in indeed a number.
Savionf

0

Your question is not clear. Why don't you post a picture of the question?
- Horta Emi
  
  0
  
  I though i did
Horta Emi

0

I posted the picture. It’s question 1 a
Savionf

0

I revised the answer and added some comments at the end.
Horta Emi

0

Thank you. Sorry for the confusion
Savionf

0

I am glad I was able to clarify :)

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