Question

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

Answer 1

Particulation solution

$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x\Rightarrow dy=\frac{2x^2+x}{x^3+x^2+x+1}dx$

Integrating both sides w.r.t x

$y=\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$

$x=-1$ as a solution of given $eqn:x^3+x^2+x+1$

as: $(-1{)}^3+(-1{)}^2+(-1)+1=0$

$\therefore (x+1)$ is a factor.

Now, we have to integrating using partial functions

$\int \frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{(x+1)}+\frac{Bx+c}{x+1}=\frac{A(x^2+1)(Bx+c)(x+1)}{(x+1)(x^2+1)}$

$\Rightarrow 2x^2+x=A(x^2+1)(Bx+c)(x+1)$

Putting $x=-1$

$2(-1{)}^2-1=A((-1{)}^2+1)\left(B\right(-1)+c)(-1+1)$

$2-1=A\left(2\right)+(-B+c)\left(0\right)$

$1=2A\Rightarrow \frac{1}{2}$

Putting $x=1$

$2\left(1\right)+1=A(12+1)+\left(B\right(1)+c)(1+1)$

$3=2A+2B+2c$

$3=3\times \frac{1}{2}+2B+2\left(\frac{-1}{2}\right)$

$3=2B$

$B=\frac{3}{2}$

Putting $x=0$

$0=A(0+1)+\left(B\right(0)+c)(0+1)$

$0=A+c\left(1\right)\Rightarrow A=-c\Rightarrow C=\frac{-1}{2}$

So, the partial equation become

$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{1}{2(x+1)}=\frac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}=\frac{1}{2(x+1)}+\frac{3x-1}{2(x^2+1)}$

Now main equation become

$y=\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$

$y=\int \frac{1}{2(x+1)}dx+\int \frac{3x}{2(x^2+1)}dx-\int \frac{1}{2(x^2+1)}dx$

$y=\frac{1}{2}\log (x+1)+\int \frac{3x}{2(x^1+1)}dx-\frac{1}{2}{\tan }^{-1}x$...(1)

Integrating $\int \frac{3x}{2(x^2+1)}dx$

Put $x^1+1=z$

$2x=\frac{dz}{dx}$

$\Rightarrow dx=\frac{dz}{zx}$

$\int \frac{3x}{2(x^2+1)}dx=\frac{3}{2}\int \frac{x}{z}x\frac{dz}{zx}=\int \frac{3}{4}\frac{dz}{z}=\frac{3}{4}\log |z|+c$d

Substituting value of $z$ we get,

$\int \frac{3x}{2(x^2+1)}dx=\frac{3}{4}\log (x^2+1)+c$

Now, from(1)

$y=\frac{1}{2}\log (x+1)+\int \frac{3x}{2(x^2+1)}dx\frac{-1}{2}ta{n}^{-1}x$

$y=\frac{1}{2}\log (x+1)+\frac{3}{4}\log (x^2+1)-\frac{-1}{2}ta{n}^{-1}+c$

Put $x=0$ and $y=1$

$1=\frac{1}{2}\log (0+1)+\frac{3}{4}\log (0+1)\frac{-1}{2}ta{n}^{-1}\left(0\right)+c$

$1=\frac{1}{2}\log \left(1\right)+\frac{3}{4}\log \left(1\right)\frac{-1}{2}ta{n}^{-1}+c$

$1=0+0-0+c$

$1=c$ $(\therefore \log 1=0andtan=0)$

Putting value of $c$ in (1)

$y=\frac{1}{2}\log (x+1)+\frac{3}{4}\log (x^2+1)\frac{1}{2}ta{n}^{-1}x+1$

$y=\frac{2}{4}\log (x+1)+\frac{3}{4}\log (x^2+1)\frac{-1}{2}ta{n}^{-1}x+1$

$y=\frac{1}{4}\log (x+1{)}^2+\frac{1}{4}\log (x^2+1{)}^3\frac{-1}{2}ta{n}^{-1}x+1$

$\therefore$ $(a\log x=log{x}^a)$

$y=\frac{1}{2}\left[\log \right(X+1{)}^2+\log (X^2+1{)}^3\frac{-1}{2}ta{n}^{-1}x+1$

$\Rightarrow y=\frac{1}{4}\log \left[\right(x+1{)}^2(x^2+1{)}^3]\frac{-1}{2}ta{n}^{-1}x+1$

$(\therefore \log a+\log b=\log ab)$