Reid's question at Yahoo Answers regarding a first order separable IVP

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SUMMARY

The particular solution to the initial value problem (IVP) defined by the equation y'x - 1 = 0 with the initial condition y(1) = 2 is y(x) = ln|x| + 2. The solution is derived by separating variables and integrating, leading to the integral form ∫2^y(x) du = ∫1^x (1/v) dv. The application of the Fundamental Theorem of Calculus (FTOC) confirms the solution through the anti-derivative process.

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MarkFL
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Here is the question:

What is the particular solution to y'x-1=0 using the inition condition x=1,y=2?

I have posted a link there to this topic so the OP can see my work.
 
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Hello reid,

We are given the IVP:

$$y'x-1=0$$ where $$y(1)=2$$

Separating the variables, we obtain:

$$dy=\frac{1}{x}\,dx$$ where $$x\ne0$$

Switching the dummy variables of integration and utilizing the boundaries as limits, we obtain:

$$\int_2^{y(x)}\,du=\int_1^x\frac{dv}{v}$$

Applying the anti-derivative form of the FTOC, we find:

$$y(x)-2=\ln|x|$$

$$y(x)=\ln|x|+2$$
 

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