The differential equations, find a particular solution satisfying the given condition: `(x^3+x^2+x+1)(dy)/(dx)=2x^2+x ; y=1`when `x = 0`

Options

A. `y=(1)/(6)[log{(x-1)^(2)(x^(2)-1)^(3)}+(1)/(2)tan^(-1)x+1`
B. `y=(1)/(4)[log{(x-1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+1`
C. `y=(1)/(4)[log{(x+1)^(2)(x^(2)-1)^(3)}]-(1)/(2)tan^(-1)x+1`
D. `y=(1)/(4)[log{(x+1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+1`

Answer

D

Explanation

Given, `(x^(3)+x^(2)+x+1)(dy)/(dx)=2x^(2)+x`
`implies dy=(2x^(2)+x)/((x^(3)+x^(2)+x+1))dx`
`implies intdy=int(2x^(2)+x)/(x^(2)(x+1)+1(x+1))dx`
`implies intdy=int(2x^(2)+x)/((x+1)(x^(2)+1)))dx`
`implies y=int(2x^(2)+x)/((x+1)(x^(2)+1))dx`………..`(1)`
Let `(ex^(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`
`implies 2x^(2)+x=x^(2)(A+B)+x(B+C)+(A+C)`
Comparing constant terms, coefficients of `x` and `x^(2)` on both sides.
`A+B=2`, `B+C=1` and `A+C=0`
Solving , we get, `A=(1)/(2)`, `B=(3)/(2)` and `C=-(1)/(2)`
put these values of `A`, `B` and `C` in equation `(2)`
`(2x^(2)+x)/((x+1)(x^(2)+1))=(1)/(2(x+1))+((3)/(2)x-(1)/(2))/(x^(2)+1)`
`implies 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`
Then from equation `(1)`,
`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)log|x+1|+(3)/(2)int(x)/((x^(2)+1))dx-(1)/(2)tan^(-1)x`
Let `x^(2)+1=timplies2x=(dt)/(dx)impliesdx=(dt)/(2x)`
`:. 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`
`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`
`implies y=(1)/(4)[2log|x+1|+3log|x^(2)+1|]-(1)/(2)tan^(-1)x+C`
`implies y=(1)/(4)[log{(x+1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+C`...........`(3)`
when `x=0`, then `y=1`, put these in equation `(3)`,
`1(1)/(4)log(1)-(1)/(2)tan^(-1)(0)+C`
`implies 1=(1)/(4)xx0-(1)/(2)xx0+CimpliesC=1`
put `C=1`, in equation `(3)` ,
`y=(1)/(4)[log{(x+1)^(2)(x^(2)+1)^(3)}]-(1)/(2)tan^(-1)x+1`
which is the required particular solution.