Solve $(1+x)y’=y$ by power series Solve $(1+x)y’=y$ by power series. $$$$Start with $y$ and $y’$:
$y=a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n$
$y’=a_1+2a_2x+3a_3x^2+...+na_nx^{n-1}$
Then $(1+x)y’=a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+...+na_n(1+x)x^{n-1}$
Then $(1+x)y’-y=[a_1(1+x)+2a_2(1+x)x+3a_3(1+x)x^2+...+na_n(1+x)x^{n-1}]-[a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n]=0$ $$$$$\implies [a_1+a_1x+2a_2x+2a_2x^2+3a_3x^2+3a_3x^3+...+(na_n+na_nx^n)]-[a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n]=0$$$$$$(a_1-a_0)+(a_1+2a+2-a_1)x+(2a_2+3a_3-a_2)x^2+(3a_3+a_3)x^3+...+=0$
I’m missing the trick when equating terms. Any help would be appreciated!
 A: Are you required to "use series"? Here's a very quick solution. Rewrite it as
$$ \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. $$
Therefore $y=c(1+x)$. 
A: Hint
1) By power serie means infinite series 
$$y=\sum_{n=0}^{\infty}a_nx^n$$
note that :
$$y'=\sum_{n=0}^{\infty}a_nnx^{n-1}=\sum_{n=1}^{\infty}a_nnx^{n-1}=\sum_{n=0}^{\infty}a_{n+1}(n+1)x^{n}$$
2) Try to use the sigma notation it's easier
$$(1+x)y’=y$$
Becomes
$$(1+x)\sum_{n=0}^{\infty}a_{n+1}(n+1)x^{n}=\sum_{n=0}^{\infty}a_nx^{n}$$
$$......$$
3) All the coefficents should be zero since we want the equality to hold $\forall \, x$
$$\sum_{n=0}^{\infty}a_nx^{n}=0 \implies a_n=0 \,\, n=1,2,3.....$$
A: Let $y(x) = \sum_{i=0}^\infty y_i x^i$.  Then $y'(x) = \sum_{i=0}^\infty i y_i x^{i-1} = \sum_{j=0}^\infty (j+1) y_{j+1} x^{j}$ where we get this by substituting $i=j+1$ and noting that when $i=0$ the term is $0$ so we don't get in trouble concerning $y_{-1}$.
Then $x y'(x) = \sum_{i=0}^\infty (i+1) y_{i+1} x^{i+1} = \sum_{j=1}^\infty j y_{j} x^{j} = \sum_{j=0}^\infty j y_{j} x^{j} - y_0 x^0$
and our equation becomes  (using the notation that $[j=0]$ means $1$ if $j$ is $0$ and $0$ otherwise)
$$
\sum_{j=0}^\infty (j+1) y_{j+1} x^{j} +  \sum_{j=0}^\infty j y_{j} x^{j} -[j=0]y_0  = \sum_{j=1}^\infty y_j x^j  
$$
Now the trick of the power series method: You can set the coefficients of each $x^j$ to match, for each individual $j$.  Thus:
$$ \forall j : (j+1)y_{j+1} + jy_j  -[j=0]y_0 = y_j 
\\ y_{j+1} = \left(-\frac{j-1- [j=0]}{j+1} \right) y_j
\\y_1 = -\frac{0-1-1}{1}y_0 = y_0
\\y_2 = -\frac{1-1-0}{2}y_1 = 0
\\y_3 = -\frac{2-1-0}{3}y_2 = -\frac23 y_2 = 0
\\ \forall j\geq 2 : y_j = 0
$$
So the general form of solution is 
$$
y = y_0 + y_0 x = y_0 (1+x)
$$
Note that because of the pleasant termination once $y_2 = 0$, this method produces an exact answer for this equation.  Also note that our excursion through the "land of log" has been eliminated.
