- #1

- 1,752

- 143

[tex]x^2 \frac{{dy}}{{dx}} + xy = 1,\,\,y\left( 1 \right) = 2[/tex]

Divide through by x^2 to set it up in a form for using the integrating factor method.

[tex]

\frac{{dy}}{{dx}} + \frac{y}{x} = - \frac{1}{{x^2 }}

[/tex]

I'm not sure where the minus sign came from on the right side. I just copied the problem from the board.

Seperate the functions of x to make it in I.F. form

[tex]

\frac{{dy}}{{dx}} + y\left( x \right)\frac{1}{x} = - \frac{1}{{x^2 }}

[/tex]

Compute the Integrating Factor with from the term that doesn't contain a y. Is this correct? Is this why 1/x was chosen to integrate and raise e to the power of it?

[tex]

I.F.\,\,e^{\int_{}^{} {\frac{1}{x}dx} } = e^{\ln x} = x

[/tex]

The next step in the notes is

[tex]

x\frac{{dy}}{{dx}} + y = \frac{1}{x}

[/tex]

Is this simply from multiplying the Integrating Factor, in this case, x, by all 3 terms in the 2nd equation? If so, what happened to my minus sign (or did it never exist and I inadvertantly put it in by mistake)?

The problem is longer than this, but I want to understand it up to this point for the moment. Thanks!