Given the IVP: $\space y'(x)=y^2-x , y(0)=1$
I need to prove using the comparison theorem that for all $x\in[0,1): 1+x\le y(x)\le \frac{1}{1-x}$
The comparison theorem I'm referring to is this one: here
My try:
For RHS: We define the following IVP $y_1'(x)=y_1^2 , y_1(0)=1$. We have $y_1(x)= \frac{1}{1-x}$ and $y'(x)\le y_1'(x) \space \forall x\in[0,1)$. Thus, we can apply the comparison theorem to get $\space y(x)\le \frac{1}{1-x}$.
Would appreciate any help or direction to prove LHS.
Probably not the way you are expected to solve this: treat it as Riccati equation, set $y=-\frac{u'}{u}$, then the associated second order linear DE is $$ u''-xu=0,~~ u(0)=1,~~u'(0)=-1 $$ The power series coefficients obey $n(n-1)a_n=a_{n-3}$, $a_0=1$, $a_1=-1$, $a_2=0$, so that $$ u(x)=1-x+\frac16x^3-\frac1{12}x^4+\frac1{180}x^6-\frac1{504}x^7\pm $$ is an alternating series with rapidly falling coefficients for $x\le 1$. This results in the bounds by partial sums $$ 1-x\le u(x) \le 1\\ 1-\frac12x^2\le -u'(x)\le 1 $$ so that combining the bounds $$ \frac{1-\frac12x^2}{1-x+\frac16x^3}\le y(x) \le \frac1{1-x}. $$ Now grind down on the left side for $x\in(0,1]$ $$ 1+x\le\frac{1-\frac12x^2}{1-x+\frac16x^3} \\ \iff 1-x^2+\frac16(x^3+x^4)\le 1-\frac12x^2 \\ \iff x+x^2\le 3 $$ and as the last inequality is true on the interval, so is the first.