85. xdydx+y−x+xycotx=0(x≠0)

85. $x \frac{d y}{d x} + y - x + x y c o t x = 0 (x \neq 0)$

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Vishal Baghel | Contributor-Level 10

8 months ago

The given D.E is

$= \frac{d y}{d x} + y \frac{(1 + x c o t x)}{x} = 1$ Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = \frac{(1 + x c o t x)}{x} & Q = 1$

$\int P d x = \int (\frac{1}{x} + \frac{x c o t x}{x}) d x = l o g | x | + l o g | s i n x | = l o g | x s i n x |$

$∴ I . F = e^{\int P d x} - e^{\int l o g | x s i n x |} x s i n x$

Thus the solution is of the form.

$\begin{array}{l} y * x s i n x = \int 1 . x s i n x d x + c \\ = x \int s i n x - \int \frac{d}{d x} \int s i n x d x d x + c \\ = - 2 c o s x + \int c o s x d x + c \\ = y * x s i n x = - x c o s x + s i n x + c \\ = y = \frac{- x c o s x}{s i n x} + \frac{s i n x}{x s i n x} + \frac{c}{x s i n x} \\ = y = - c o t x + \frac{1}{x} + \frac{c}{x s i n x} \end{array}$

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dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.