Question

For the differential equation in given question find a particular solution satisfying the given condition.

$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x,y=1whenx=0.$

Answer 1

Given, $(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x$
On separating the variables, we get $dy=\frac{2x^2+x}{(x^2+x^2+x+1)}dx$
On integarting both sides, we get
$\int dy=\frac{2x^2+x}{(x^3+x^2+x+1)}dx\Rightarrow \int dy=\int \frac{2x^2+x}{x^2(x+1)+1(x+1)}dx\Rightarrow \int dy=\int \frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{(x+1)}+\frac{Bx+C}{(x^2+1)}\left[bypartialfraction\right]....\left(i\right)\Rightarrow \frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+1)}\Rightarrow 2x^2+x=A{x}^2+A+B{x}^2+Bx+Cx+C\Rightarrow 2x^2+x=x^2(A+B)+x(B+C)+(A+C)$
On comparing the coefficients of $x^2,$x and constant terms on both sides,
we get
A+B=2, B+C=1 and A+C=0
On solving these equations, we gte $A=\frac{1}{2},b=\frac{3}{2}andC=-\frac{1}{2}$
On substituting the values of A, B and C in Eq. (ii), we get
$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{1}{2(x+1)}+\frac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}$
$\therefore \int \frac{2x^2+x}{(x^2+1)(x+1)}dx=\frac{1}{2}\int \frac{1}{(x+1)}dx+\int \frac{\frac{3}{2}x-\frac{1}{2}}{(x^2+1)}dx$
Then from Eq.(i), we get
$y=\frac{1}{2}log|x+1|+\frac{3}{2}\int \frac{x}{(x^2+1)}dx-\frac{1}{2}\int \frac{1}{(x^2+1)}dx\Rightarrow y=\frac{1}{2}log|x+1|+\frac{3}{2}\int \frac{x}{(x^2+1)}dx-\frac{1}{2}ta{n}^{-1}xLet{x}^2+1=t\Rightarrow 2x=\frac{dt}{dx}=dx=\frac{dt}{2x}\therefore y=\frac{1}{2}log|x+1|+\frac{3}{2}\int \frac{x}{t}\frac{dt}{2x}-\frac{1}{2}ta{n}^{-1}x\Rightarrow y=\frac{1}{2}log|x+1|+\frac{3}{4}\int \frac{1}{t}dt-\frac{1}{2}ta{n}^{-1}x$

$\Rightarrow y=\frac{1}{2}log|x+1|+\frac{3}{4}log|t|-\frac{1}{2}ta{n}^{-1}x+C\Rightarrow y=\frac{1}{2}log|x+1|+\frac{3}{4}log|x^2+1|-\frac{1}{2}ta{n}^{-1}x+C\Rightarrow y=\frac{1}{4}[2log|x+1|+3log|x^2+1|]-\frac{1}{2}ta{n}^{-1}x+C$
$\Rightarrow y=\frac{1}{4}\left[log\right((x+1{)}^2(x^2+1{)}^3\left)\right]-\frac{1}{2}ta{n}^{-1}x+C$ ..... (iii)
$[\because log{m}^n=nlogm]$
When x=0, then y = 1 on putting these values in Eq. (iii), we get
$1=\frac{1}{4}log\left(1\right)-\frac{1}{2}ta{n}^{-1}\left(0\right)+C\Rightarrow 1=\frac{1}{4}\times 0-\frac{1}{2}\times 0+C\Rightarrow C=1$
On putting C = 1 in Eq. (iii), we get
$y=\frac{1}{4}\left[log\right((x+1{)}^2(x^2+1{)}^3\left)\right]-\frac{1}{2}ta{n}^{-1}x+1$
which is the required particular solution.