अवकल समीकरण को हल कीजिए-
`x log x (dy)/(dx) +y =2/x log x.`

Explanation

दिया गया अवकल समीकरण है-
`x log x (dy)/(dx) +y=2/x log x`
`implies(dy)/(dx) +((1)/(x log x)) y=(2 log x)/(x^(2)log x)`
`implies(dy)/(dx)+((1)/(x log x))y= (2)/(x^(2))" "...(1)`
जो कि y रैखिक अवकल समीकरण है.
समी (1 ) की तुलना मानक रूप `(dy)/(dx) +Py =Q` से करने पर,
`P =(1)/(xlog x ) ` और `Q =(2)/(x^(2))`
`thereforeI .F. =e^(int Pdx)=e ^(int (1)/(x log x))`
`e ^(int (1)/(t)dt) , [" माना" log x =1 implies(1)/(x)dx =dt ]`
`=e ^(log |t|)=|t|=log x`
अतः अभीष्ट हल है-
`yxx (I.F.) =int Q . (I. F.) dx+C`
`impliesy log x = int (2)/(underset(II)(x^(2)))underset(I)(log ) x dx +C`
`implies y log x = log x int (2)/(x^(2)) dx`
`-int {(d)/(dx)(log x). int (2)/(x^(2))dx}dx+C`
`impliesy log x = log x (-(2)/(x))-int ((1)/(pi)xx(-2)/(x))dx+C`
`impliesy log x =- (2)/(x) log x+ int (2)/(x^(2))dx+C`
`implies y log x =-(2)/(x) log x- (2)/(x)+C.`