`(dy)/(dx)=2x^(2)+x,y=1` यदि x=0

Explanation

दिया है, `(x^(3)+x^(2)+x+1)(dy)/(dx)=2x^(2)+x`
`implies dy=(2x^(2)+x)/((x^(3)+x^(2)+x+1))dx`
`implies int dy = int(2x^(2)+x)/(x^(2)(x+1)+1(x+1))dx`
`= int dy = int (2x^(2)+x)/((x+1)(x^(2)+1))dx`
`implies y=int(2x^(2)+x)/((x+1)(x^(2)+1))dx" ....(1)"`
माना `(2x^(2)+x)/((x+1)(x^(2)+1))=(A)/((x+1))+(Bx+C)/((x^(2)+1))" ....(2)"`
`implies (2x^(2)+x)/((x+1)(x^(2)+1))=(A(x^(2)+1)+(Bx+C)(x+1))/((x+1)(x^(2)+1))`
`implies 2x^(2)+x=Ax^(2)+A+Bx^(2)+Bx+Cx+C`
`.^(____)2x^(2)+x=x^(2)(A+B)+x(B+C)+(A+C)`
दोनों और अचर पद, x तथा `x^(2)` के गुणांकों की तुलना करने पर,
`A+B=2,B+C=1` तथा `A+C=0`
समीकरणों को हल करने पर, `A=(1)/(2), B=(3)/(2)` तथा `C=-(1)/(2)` A, B तथा C के मान समीकरण (2) में रखने पर,
`(2x^(2)+x)/((x+1)(x^(2)+1))=(1)/(2(x+1))+((3)/(2)x-(1)/(2))/(x^(2)+1)`
`=int(2x^(2)+x)/((x^(2)+1)(x+1))dx`
`=(1)/(2)int(1)/((x+1))dx+int((3)/(2)x-(1)/(2))/((x^(2)+1))dx`
तब, समीकरण (1) से,
`y=(1)/(2)log|x+1|+(3)/(2)int(x)/((x^(2)+1))dx-(1)/(2)int(1)/((x^(2)+1))dx`
`implies y=(1)/(2)log|x+1|+(3)/(2)int(x)/((x^(2)+1))dx-(1)/(2)tan^(-1)x`
माना `x^(2)+1=timplies2x=(dt)/(dx)impliesdx=(dt)/(2x)`
`therefore y=(1)/(2)log|x+1|+(3)/(2)int(x)/(t)(dt)/(2x)-(1)/(2)tan^(-1)x`
`implies y=(1)/(2)log|x+1|+(3)/(4)int(1)/(t)dt-(1)/(2)tan^(-1)x`
`implies y=(1)/(2)log|x+1|+(3)/(4)log|t|-(1)/(2)tan^(-1)x+C`
`implies y=(1)/(2)log|x+1|+(3)/(4)log|x^(2)+1|-(1)/(2)tan^(-1)x+C`
`impliesy=(1)/(4)[2log|x+1|+3log|x^(2)+1|]-(1)/(2)tan^(-1)x+C`
`impliesy=(1)/(4)[log{(x+1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+C" ....(3)"`
जब x = 0, तब y = 1 समीकरण (3) में रखने पर,
`1=(1)/(4)log(1)-(1)/(2)tan^(-1)(0)+C`
`implies1=(1)/(4)xx0-(1)/(2)xx0+CimpliesC=1`
C = 1, समीकरण (3) में रखने पर,
`y=(1)/(4)[log{(x+1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+1`
जोकि अभीष्ट व्यापक हल है।