Question

If $y^2+2y=x(y+1)$, show that one value of $y$ is
$\frac{1}{2}x+\frac{1}{8}{x}^2-\frac{1}{128}{x}^4+\cdots$.

Answer 1

To show that one value of $y$ is

$\frac{1}{2}x+\frac{1}{8}{x}^2-\frac{1}{128}{x}^4+.....$

Since, $y=0$ when $x=0$

We may assume $y=A_{1}x+A_2{x}^2+A_3{x}^3+.......$ ;

Substitute this value for $y$ in the given relation, then

${(A_{1}x+A_2{x}^2+A_3{x}^3+...)}^2+2(A_{1}x+A_2{x}^2+A_3{x}^3+....)=x(A_{1}x+A_2{x}^2+A_3{x}^3+...+1)$

Since this is an identity, we may equate the coefficients of power of $x$ and thus we obtain

$2A_1=1$ or $A_1=\frac{1}{2}$

$A_1^2+2A_2=A_1$, whence $A_2=\frac{1}{8}$

$2A_1{A}_2+2A_3=A_2$, whence $A_3=0$

$A_2^2+2A_1{A}_3+2A_4=A_3$

$\Rightarrow A_4=\frac{-1}{128}$

$\therefore$ $y=\frac{1}{2}x+\frac{1}{8}{x}^2-\frac{1}{128}{x}^4+....$