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[crude yoke]

Is this result true?

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[inland copper]

I think you forgot to multiply the -2(x - 2)^2 term when moving it to the RHS.
Anyway, seems mostly ok apart from that, though you do have some strange expressions at the end. Here's an quicker way to do it.
(x - 2)y' - y = 2(x - 2)^3
((x - 2)y' - y)/(x - 2)^2 = 2(x - 2)
(y/(x - 2))' = 2(x - 2)
Then integrate and express y.

[crude yoke]

This is actually what my friend sent me i did it in this way but im stuck

[crude yoke]

Can u please tell me what i did is right?

[inland copper]

Oh, the general formula. Was it like this, though? Let me check.

[crude yoke]

U can do it in two ways find the integrating factor and mulitiply the eq1 with it or use the general form

[inland copper]

Those aren't the only ways. At least, you can write it in another way, even without using the shortcut for this case that I did above.

[crude yoke]

Im stuck at that integral step

[crude yoke]

So what i did is right also?

[inland copper]

This is the general formula for y' + f(x)y = g(x). Note that the integrals are taken as antiderivatives, so without C.
We have:
f(x) = -1/(x - 2)
g(x) = 2(x - 2)
∫(f(x)dx) = -ln(|x - 2|)
e^(-∫(f(x)dx)) = x - 2
g(x)e^(∫(f(x)dx)) = 2(x - 2)/(x - 2) = 2
∫(g(x)e^(∫(f(x)dx))dx) = 2x
So:
y = (x - 2)(C + 2x)

[inland copper]

I'm not sure what expressions you have near the end. The result is correct, so just try writing it a bit more clearly.

As for what general approach I prefer: solving the homogeneous equation, then using variation of parameters.

[crude yoke]

Like the one my friend did?

[inland copper]

Well, your friend just used the general formula.

It will come from any approach, but it might not be easy to remember.

[crude yoke]

Thank you so much I appreciate it

[civic ice]

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