Question

ds

Answer 1

\[\frac{\text dx}{\text dy}=\frac{y^3-x^3\sin(y/x)}{xy^2}\]

Answer 2

isn't is dy/dx equals that stuff?

Answer 3

i was just copying the question

Answer 4

yeah but the first part should be dy/dx

Answer 5

to get this equation in homogenous form
first break up the fraction, into a difference of fraction , and do some canceling
\[\frac{\text dx}{\text dy}=\frac{y^3-x^3\sin(y/x)}{xy^2}\]

Answer 6

\[\frac{\text dx}{\text dy}=\frac{y^3}{xy^2}-\frac{x^3\sin(y/x)}{xy^2}=\quad \dots\]

Answer 7

dy/dx = (y/x) -((x^2sin(y/x))/y^2)

Answer 8

when this have been done
substitute \(y=vx\) or simply \(\frac yx =v\)
\[\frac{\text dy}{\text dx}=v+xv'\]

Answer 9

v + xv' = v-((x^2sin(v))/y^2)

Answer 10

@UnkleRhaukus then what?

Answer 11

\[\frac{\text dx}{\text dy}=\frac{y}{x}-\frac{x^2}{y^2}\sin(y/x)\]\[\qquad\qquad=\frac{y}{x}-\left(\frac{x}{y}\right)^2\sin\left(\frac{y}{x}\right)\]

now do the substitution

Answer 12

Why do you say dx/dy is that? isnt it dy/dx

Answer 13

oh yes my mistake

Answer 14

\[v+xv'=v-\frac{\sin(v)}{v^{2}}\]

Answer 15

\[xv'=-\frac{\sin(v)}{v^{2}}\]\[xv^2v'=-\sin(v)\]\[xv^2v'+\sin(v)=0\]