Solve $(1+x)y’=y$ by power series. $$$$Start with $y$ and $y’$:
$y=a_0+a_1x+a_2x^2+a_3x^3+…+a_nx^n$
$y’=a_1+2a_2x+3a_3x^2+…+na_nx^{n-1}$
Then $(1+x)y’=a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+…+na_n(1+x)x^{n-1}$
Then $(1+x)y’-y=[a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+…+na_n(1+x)x^{n-1}]-[a_0+a_1x+a_2x^2+a_3x^3+…+a_nx^n]=0$ $$$$$\implies [a_1+a_1x+2a_2x+2a_2x^2+3a_3x^2+3a_3x^3+…+(na_n+na_nx^n)]-[a_0+a_1x+a_2x^2+a_3x^3+…+a_nx^n]=0$$$$$$(a_1-a_0)+(a_1+2a+2-a_1)x+(2a_2+3a_3-a_2)x^2+(3a_3+a_3)x^3+…+=0$
I’m missing the trick when equating terms. Any help would be appreciated!
Best Answer
Are you required to "use series"? Here's a very quick solution. Rewrite it as $$ \frac{y'}{y} = \frac{1}{1+x}. $$ Note that $$ \frac{y'}{y} = \frac{d\phantom{x}}{dx}\ln(y) $$ and $$ \frac{1}{1+x} = \frac{d\phantom{x}}{dx}\ln(1+x). $$ Integrating gives $$ \ln(y) = \ln(1+x) + C. $$ Therefore $y=c(1+x)$.