अवकलन समीकरण हल कीजिए :
`(1+ x^(2)) dy + 2xydx = cot x dx `

Explanation

दिया गया अवकल समीकरण है-
`(1+x^(2))dt+2xydx=cot xdx`
`implies(1+ x^(2))(dy)/(dx) +2xy =cot `
`implies(dy)/(dx) +(2x)/(1+x^(2))y=(cot x)/(1+c^(2))" "...(1)`
यहाँ `P=(2x)/(1+x^(2)),Q=(cot x)/(1+x^(2))`
` therefore I . F. =e ^(intPdx)=e ^(int (2x)/(1+x^(2))dt)`
`=e^(int (dt)/(t)), ["माना"1+x^(2)=timplies 2 xdx=dt]`
`=e^(log|t|)=e^(log |1+x^(2)|)=1+x^(2)`
यही अभीष्ट व्यापक हल है-
`yxx (I.F.) =int Q xx (I.F.)dt +C`
`impliesy(1+x^(2))=int(cot x)/(1+x^(2))xx(1+x^(2)) dx+C`
`impliesy(1+x^(2))=intcot x dx +C`
`impliesy (1+x^(2))=log |sin x| +C`
`implies y=(log |sin x|)/(1+x^(2))+(C)/(1+x^(2)).`