Playing With Logs while solving y-x=5(ln|y+3|-ln|x+4|) for y.

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In summary, the conversation discusses a problem involving a separable differential equation and the techniques used to solve it, including factorization, integration, and logarithmic identities. The resulting equation is y-x=5(ln|y+3|-ln|x+4|), but there is no apparent way to solve for y in terms of x.
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gikiian
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Homework Statement



The problem is actually the following Separable Differential Equation:
[itex]\frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}[/itex]​
I am required to find y(x).2. Homework Equations and techniques

- factorization (applied on the numerator and the denominator in the problem equation)
- basic integration (applied after separating the variables)
- at least the following logarithmic identities:
[itex]e^{ln|x|}=x[/itex] ; [itex]ln|xy|=ln|x|+ln|y|[/itex]​

The Attempt at a Solution



[itex]\frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}[/itex]

[itex](1-\frac{5}{y+3})dy=(1+\frac{5}{x+4})dy[/itex]

[itex]y-5ln|y+3|=x+5ln|x+4|+C[/itex]

[itex]y-x=5(ln|y+3|-ln|x+4|)[/itex]


I also put this result in Wolfram Alpha, but it could not solve the equation for y. My instructor apparently believes that an y=f(x) can be obtained. Is there any way to do so?
 
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  • #2
You appear to have a sign wrong in the middle two equations, but since it comes out right at the end maybe this was a copying out error.
i don't see a way to get it into y = f(x) form either.
 
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1. What is the purpose of using logs while solving this equation?

The purpose of using logs is to make the equation easier to solve by converting exponential expressions into simpler forms.

2. How do I determine which base to use when using logs?

The base of the log should match the base of the exponential expression in the equation. In this case, the base would be e (natural logarithm) since the equation contains ln.

3. Can I use any log to solve this equation?

No, it is important to use the same base for both sides of the equation when solving with logs. If different bases are used, the solution will not be accurate.

4. How do I solve for y in this equation?

To solve for y, you will need to use the properties of logarithms to simplify the equation. Then, you can solve for y by isolating it on one side of the equation.

5. Are there any restrictions or special cases when solving this equation with logs?

Yes, since the equation contains logs, the values inside the logarithms must be positive. This means that the expressions within the absolute value signs (|y+3| and |x+4|) must be greater than 0.

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