$ cot (y/x) = log |ex| $
· A differential equation of the form $dy/dx = F(x,y)$ is said to be homogenous if $F(x,y)$ is a homogenous function of degree zero.
· To solve equations of these types, we make substitution $y = vx$ differentiating with respect to $x$ we get, $dy/dx = v + xdv/dx$
· If the homogenous differential function is in the form $dx/dy= F(x,y)$ where $F(x,y)$ is homogenous function of degree zero, then we make substitution $x = vy$, then on differentiating we get $dx/dy = v + ydv/dy.$
Let us find out whether the given problem is a homogenous solution with the degree zero or not.
Let us rewrite the given problem as $dy/dx = -[xsin^2(x/y) - y]/x$
i.e., $F(x,y) = -[xsin^2(x/y) -y]/x$
$F(kx.ky) = -[kxsin^2(kx/ky) - ky]/kx = - [xsin^2(x/y) - y)]/x = k^0.F(x,y)$
Hence it is a differential equation with degree 0.
Hence we substitute $y = vx$ and $dy/dx = v + xdv/dx$
Therefore $v+xdv/dx = -[xsin^2v - vx]/x$.
Cancelling the common factor $x$ we get,
$ v + xdv/dx = -[sin^2v-v] = v - sin^2v$.
On cancelling v we get $dv/dx = - sin^2v$, bringing $dv$ and $v$ on the LHS and $x$ and $dx$ on the RHS, we get
$dv/sin^2v = -dx/x$, but $1/sin^2x = cosec^2x$
Therefore $cosec^2v dv = - dx/x$
Integrating on both sides $-cotv = - log|x| + logC$
Now substituting for $v$ as $y/x$ we get $cot(y/x) = logcx$
Given $y = ¶/4$ and $x = 1$, now substituting for $x$ and $y$ we get $cot(¶/4) = log|Cx1|$, we know cot $¶/4 = 1$
Hence $1 = logc$ or $c = e^1$ or $c = e$
Now substituting for C we get the solution as $cot(y/x) = log|ex|$