Question

e^ydx-x(2xy+e^y)dy=0

Answer 1

\(e^ydx-x(2xy+e^y)dy=0\)

exact or not?

Answer 2

nah

Answer 3

its not

Answer 4

you should learn to typeset.

Answer 5

is anyone in here in a k12 school

Answer 6

typeset?

Answer 7

\(t^{y^p} \dfrac{e}{s} e^t \)

Answer 8

Hugh Mungus

Answer 9

\(\color{black}{\displaystyle e^ydx-(2x^2y+xe^y)dy=0}\)

\(\color{black}{\displaystyle e^y\ne4xy+e^y\quad \Longrightarrow \quad\frac{dM}{dy}\ne\frac{dN}{dx}}\)

try integrating factor.

Answer 10

One of the two:

\(\color{black}{\displaystyle g(x)=\frac{1}{N}[M_y-N_x]}\)

\(\color{black}{\displaystyle h(y)=\frac{1}{M}[N_x-M_y]}\)

Answer 11

\[e^ydx=(2x^2y+x e^y)dy\]
\[e^y \frac{ dx }{ dy }=2x^2y+xe^y,\frac{ dx }{ dy }-x=2x^2y e ^{-y}\]
divide by x^2
\[x^{-2}\frac{ dx }{ dy }-x^{-1}=2ye^{-y}\]
Put~\[Put~x^{-1}=t\]
\[-x^{-2}\frac{ dx }{ dy }=\frac{ dt }{ dy }\]
\[-\frac{ dt }{ dy }-t=2ye^{-y},\frac{ dt }{ dy }+t=-2ye^{-y}\]

Answer 12

\[I.F=e^{\int\limits 1dy}=e^y\]
c.s. is
\[t.e^y=-2 \int\limits ye^{-y}e^ydy=-2 \int\limits ydy+c=-2*\frac{ y^2 }{ 2 }+c=-y^2+c\]
replace the value of t and get the result.