1. Originally Posted by Prove It
Step 1: Solve for $\displaystyle y^{-2}$.

Step 2: Take both sides to the power of [tex]\displaystyle -\frac{1}{2}[/math].
i have never seen this before... i would only have thought of multiplying by y^3 but that wouldnt help

what do you do with a -1/2 power? 1/2 is root but what do you do with minus half? and do you apply it to thE WHOLE rhs or just the individual terms>?

ah, is it (1/y^2)^-1/2 = 1/(1/(sqrt(y^2)) = 1/(1/y) = y

same thing to the rhs?

so it becomes 1/(sqrt(rhs))?

2. Originally Posted by Prove It
Except $\displaystyle \frac{dv}{dx} - \frac{3v}{x} = \frac{x + 1}{x}$ is not a homogeneous equation, and is not separable.

Look to my answer #6. I said solve the homogeneous equation. For a lineal equation always the homogeneous is separable. Solving this one, you can use (obviously that was my proposal) the variation of constants method.

Since it's first-order linear, the Integrating Factor method is needed...

It is not necessary.

Fernando Revilla

3. it worked just fine and the i.f. method is easier anyway. plus as i said, this question does not require knowledge of homogeneous equations so that was not the intended method

4. Originally Posted by mathcore
it worked just fine and the i.f. method is easier anyway. plus as i said, this question does not require knowledge of homogeneous equations so that was not the intended method

All right, but up untill my answer #6 nobody could know which was the intended method.

Fernando Revilla

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