Question

\[\text{Solve the differential equation}\]
\[\text dx+\left(\frac xy-\sin y\right)\text dy=0\]

Answer 1

another linear one in x

Answer 2

\[M=1\qquad\qquad N=\frac xy-\sin y\]\[\frac{\partial M}{\partial y}=0\qquad\qquad \frac{\partial N}{\partial x}=\frac 1y\]\[\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}\]
\[\text{let}\quad R=R(y)\]\[\frac{\partial MR(y)}{\partial y}= \frac{\partial NR(y)}{\partial x}\]\[R(y)\frac{\partial M}{\partial y}+MR'(y)=R(y)N'\]\[R'(y)=\frac{R(y)}{y}\]\[y\frac{R'(y)}{R(y)}=1\]\[\int y\frac{\text dR(y)}{R(y)}=\int 1 \text dx\]

Answer 3

the solutions say that integrating factor \(R(y)=y\)

Answer 4

I think so too ...
\[ R'(y) = R_y(y)\]

Answer 5

so you have
\[ \int \frac{dR(y)}{R(y)} = \frac{dy}{y}\]

Answer 6

\[\ln R(y)=\ln y+c\]\[R(y)=ky\]

Answer 7

IF do not have k's

Answer 8

oh ok

Answer 9

\[R'(y)=\frac{R(y)}{y}\]\[\frac{R'(y)}{R(y)}=\frac 1y\]\[\int \frac{R'(y)}{R(y)}=\int\frac 1y\]\[\ln R(y)=\ln y\]\[R(y)=y\]

Answer 10

not quite right notation
\[ \frac{dR(y)}{R(y)}=\frac {dy}y \]
\[ \int \frac{dR(y)}{R(y)}=\int\frac 1y dy\]

Answer 11

yeah thank you this looks better

\[\frac{R'(y)}{R(y)}=\frac 1y\]\[\frac{\text dR(y)}{R(y)}=\frac 1y\text dy\]\[\int \frac{\text dR(y)}{R(y)}=\int\frac 1y\text dy\]\[\ln R(y)=\ln y\]\[R(y)=y\]

Answer 12

\[\frac{\partial MR(y)}{\partial y}=\frac{\partial y}{\partial y}=1\]\[\frac{\partial NR(y)}{\partial x}=\frac{\partial (x-y\sin y)}{\partial x}=1\]\[\frac{\partial MR(y)}{\partial y}= \frac{\partial NR(y)}{\partial x}\]

Answer 13

\[f(x,y)=\int NR(y)\text dy+g(x)=\int MR(y)\text dx+h(y)\]
\[=\int NR(y)\text dy=\int x- y\sin y\text dy+g(x)\]

Answer 14

how do i integrate \( y\sin y\) ?

Answer 15

or should i be doing the \(\int MR(y)\text dy+h(x)\)

Answer 16

*\(\text dx\)

Answer 17

/?

Answer 18

a better one
http://www.wolframalpha.com/input/?i=integrate+x+-+y+siny+with+dy
make use of tech man!!

Answer 19

\[f(x,y)=\int NR(y)\text d y+g(x)\]\[=\int x -y\sin \text dy+g(x)\]\[=xy-\sin y-y\cos (y)+g(x)\]
\[MR(y)=y\]

Answer 20

\[ F(x,y) = xy−\sin y−y\cos (y)+g(x)
\]
\[ F_x(x,y) = 1 + g'(x)\\
y = 1 + g'(x)
\\
g(x) = \int (y - 1)dx\]

Answer 21

\[f(x,y)=xy−\sin y−y \cos (y)+g(x)\]
\[\frac{\partial f(x,y)}{\partial x} =y+g'(x)\]

Answer 22

Oh sorry ... y's there .. it get's canceled and, g'(x) = constant

Answer 23

\[f(x,y)=\int NR(y)\text d y+g(x)\]\[=\int x -y\sin \text dy+g(x)\]\[f(x,y)=xy+y \cos (y)-\sin y+g(x)\]\[\frac{\partial f(x,y)}{\partial x} =y+g'(x)\]\[MR(y)=y\]\[g'(x)=0\]\[g(x)=c_1\]\[f(x,y)=xy+y \cos y-\sin y+c_1\]\[f(x,y)=xy+y \cos y-\sin y=c\]

Answer 24

*id changed a + to a - buy mistake , fixed now,
and this agrees with the answerin the back of my book
Thanks for the assistance @experimentX