Answer
$y=3(3x-\tan 3x)$
Work Step by Step
We are given:
$$y'=(9x-y)^2$$
$$F(ax+by+c)=(9x-y)^2$$
\[\frac{dy}{dx}=F(ax+by+c)\;\;\;\ldots \]
Substitute $\:V=9x-y$
we have: $$\frac{dV}{dx}=9-\frac{dy}{dx} (1)$$
$$\frac{dy}{dx}=V^2 (2)$$
From (1) and (2)
\[\frac{dV}{dx}=9-V^2\]
Separating variables
\[\frac{1}{bF(V)+a}dV=dx\]
Integrating both sides:
$$\int \frac{1}{9-V^2}dV=\int dx$$
To integrate the left side:
$$V=3t \rightarrow dV=3dt$$
$$\rightarrow \int \frac{1}{9-V^2}dV=\int \frac{3}{9-9t^2}=\frac{1}{3}\int \frac{1}{1-t^2}dt=\frac{1}{3}\int \frac{1}{(1-t)(1+t)}=\int \frac{A}{1-t}+\frac{B}{1+t}dt$$
$$\rightarrow 1 = A+At+B-Bt$$
$\rightarrow A+B=1$ or $A - B=0$
$A=B$ or $2A=1$
$A=\frac{1}{2}$ or $B=\frac{1}{2}$
$$\int \frac{1}{(1-t)(1+t)}dt=\frac{1}{2}\int \frac{1}{1-t}dt+\frac{1}{2}\int \frac{1}{1+t}dt=\frac{1}{2}(-\ln(1-t)+\ln(1+t))$$
$$\int \frac{1}{9-V^2}dV=\frac{1}{6}(\ln (1+\frac{V}{3})-\ln(1-\frac{V}{3})=\frac{1}{3}\tan ^{-1}(\frac{V}{3})$$
Hence here,
$$\int \frac{1}{9-V^2}dV=\int dx$$
$$\frac{1}{3}\tan^{-1}(\frac{V}{3})=3x+C$$
$$\tan^{-1}(3x-\frac{y}{3})=3x+C$$
$$3x-\frac{y}{3}=\tan (3x+C)$$
$$y=3(3x-\tan 3x + C)$$
We are given: $y(0)=0$
$$\rightarrow 0=3(0-\tan (C))$$
$$\rightarrow C=0$$
The final solution is:
$$y=3(3x-\tan 3x)$$
