|xy| = ce^x

x + log (1 - y) = c

y = log |e^x+e^-x|+c

y = log(x²+1)+c

\(x\frac { dy }{ dx } =ylogy\)

\(x\frac { dy }{ dx } =2y\)

\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)

order 2 ; degree 1

order 2;degree not defined

order – 2 : degree 2

This is a linear differential equation
Here P = 1, Q = e^-x
\(\therefore \int { p\ dx } =\int { 1.dx } =x\)
\(I.F={ e }^{ \int { pdx } }={ e }^{ x }\)
The solution is
\({ ye }^{ \int { pdx } }\int { { Qe }^{ \int { p\ dx } }dx+c } \)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ -x }.{ e }^{ x }dx+c=\int { dx+c } } \)
\(\int { y{ e }^{ x }=x+c } \)

Given  x \(\frac{dy}{dx}=x+y\)
\(\frac{dy}{dx}=\frac{x+y}{x}\)    ...(1)
This is a homogeneous differential equation
put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes, 
\(v+x\frac { dv }{ dx } =\frac { x+vx }{ x } =1+v\)
\(\Rightarrow x\frac { dv }{ dx } =1+v-v=1\)
\(\Rightarrow dv=\frac { dx }{ x } \)
\(\Rightarrow \int { dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow v=log\quad x+c\)
\(\\ \Rightarrow \frac { y }{ x } =log\ x+c[\because v=\frac { y }{ x } ]\)

Given \(\frac{dy}{dx}=1+e^{x-y}\) ...(1)
putting x - y = z ⇒ 1 - \(\frac{dy}{dx}=\frac{dz}{dx}\)
\(\Rightarrow \frac { dy }{ dx } =1-\frac { dz }{ dx } \)
∴ (1) becomes,
\(1-\frac { dz }{ dx } =1+{ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ dx } ={ e }^{ z }\)
\(\Rightarrow -\frac { dz }{ { e }^{ z } } =dx\)
\(-\int { { e }^{ -z }dz } =\int { dx } \)
\(\Rightarrow \frac { { e }^{ -z } }{ -1 } =x+c\)
\(\Rightarrow { e }^{ y-x }=x+c\)

Given slope = \(\frac{dy}{dx}=e^x\)
\(\Rightarrow dy={ e }^{ x }dx\)
\(\int { dy } =\int { { e }^{ x }dx } \)
\(\Rightarrow y={ e }^{ x }+c\)
Since the curve passes through (0, 0),
0 = e⁰+c
⇒ 0 = 1+c
⇒c = -1
y = ex^-1 is the required equation of the curve

Equation of family of straight lines in my plane is y = mx - c where m and c are arbitrary constraints.
Differentiating, y' = m
Differentiating again, y" = 0, is the required differential equation.