Question

need help with a differentials question

Answer 1

The equation is linear in \(y\). You can find an integrating factor and proceed from there.

Answer 2

would that be e to the integral of 2/x - 1?

Answer 3

I think its this:
\[e^{\int\limits (\frac{-2}{x}-1) dx}\]

Answer 4

oh because of the negative before y gotcha thank you!

Answer 5

also i had another question why is it -1 and not +1? @freckles

Answer 6

\[xy'-2y=xy+xe^x \\ xy'-2y-xy=xe^x \\ y'-\frac{2y}{x}-\frac{xy}{x}=\frac{xe^x}{x} , x \neq 0 \\ y'+(\frac{-2}{x}-1)y=e^x\]

Answer 7

gotcha :) I was doing it using that and I got up to this part e^(-2lnx-x)y=-xe^(2lnx)/2 +C

Answer 8

how did you get your right hand side

Answer 9

Could you help me with how I should separate y, I'm not sure what happens to the C when I devide with e @freckles

Answer 10

\[(yv)'=ve^x \\ v=e^{-2\ln|x|-x}=e^{\ln(x^{-2})-x}=e^{\ln(x^{-2})}e^{-x}=x^{-2}e^{-x} \\ \\ \text{ so we have } (yx^{-2}e^{-x})'=x^{-2}e^{-x}e^x \\ yx^{-2}e^{-x}=\int\limits x^{-2} dx\]

Answer 11

This is what i did

Answer 12

I don't know how you integrate 1/x^2 like that

Answer 13

\[e^{-2 \cdot \ln(x)}=e^{\ln(x^{-2})} \text{ by power rule } \\ =x^{-2} \text{ since } y=e^x \text{ and } y=\ln(x) \text{ are inverses }\]

Answer 14

Ah that makes so much sense thank you!!