Question

For the differential equations , find a particular solution satisfying the given condition:$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x;y=1$ when $x=0$

Answer 1

The given differential equation is:
$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x$
$⇒\frac{dy}{dx}=2x^2+\frac{x}{(x^3+x^2+x+1)}$
$⇒dy=2x^2+\frac{x}{(x+1)(x^2+1)}dx$
Integration both sides,we get:
$∫dy=∫2x^2+\frac{x}{(x+1)(x^2+1)}dx...(1)$
Let $2x^2+\frac{x}{(x+1)(x^2+1)}=\frac{A}{x}+1+Bx+\frac{C}{x^2}+1...(2)$
$⇒2x^2+\frac{x}{(x+1)(x^2+1)}=Ax^2+A+\frac{(Bx+C)(x+1)}{(x+1)(x^2+1)}$
$⇒2x^2+x=Ax^2+A+Bx^2+Bx+Cx+C$
$⇒2x^2+x=(A+B)x^2+(B+C)x+(A+C)$
Comparing the coefficients of $x^2$and $x$,we get:
$A+B=2$
$B+C=1$
$A+C=0$
Solving these equations,we get:
$A=\frac{1}{2},B=\frac{3}{2},and C=\frac{-1}{2}$
Substituting the values of A,B,and C in equation(2),we get:
$2x^2+\frac{x}{(x+1)(x^2+1)}=\frac{1}{2}.\frac{1}{(x+1)}+\frac{1}{2}\frac{(3x-1)}{(x^2+1)}$
Therefore,equation(1)becomes:
$∫dy=\frac{1}{2}∫\frac{1}{x}+1 dx+\frac{1}{2}∫3x-\frac{1}{x^2}+1dx$
$⇒y=\frac{1}{2}log(x+1)+\frac{3}{2}∫\frac{x}{x^2+1}dx-\frac{1}{2}∫\frac{1}{x^2}+1dx$
$⇒y=\frac{1}{2}log(x+1)+\frac{3}{4}.∫\frac{2x}{x^2+1}dx-\frac{1}{2}tan^{-1}x+C$
$⇒y=\frac{1}{2}log(x+1)+\frac{3}{4}log(x^2+1)-\frac{1}{2}tan^{-1}x+C$
$⇒y=\frac{1}{4}[2log(x+1)+3log(x^2+1)]-\frac{1}{2}tan^{-1}x+C$
$⇒y=\frac{1}{4}[(x+1)^2(x^2+1)^3]-\frac{1}{2}tan^{-1}x+C...(3)$
Now,y=1,when x=0.
$⇒1=\frac{1}{4}log(1)-\frac{1}{2}tan^{-1}0+C$
$⇒1=\frac{1}{4}\times0-\frac{1}{2}\times0+C$
$⇒C=1$
Substituting C=1 in equation(3),we get:
$y=\frac{1}{4}[log(x+1)^2(x^2+1)^3]-\frac{1}{2}tan^{-1}x+1.$