Solving a two function differential equation I came across this question and I have tried a lot but still cannot get it.Can someone help??
$$y'(x)\phi(x)-y(x)\phi'(x)+y^2(x)=9$$
Given that $y(1)=1$.
We need to find $y(2)$.
Also we need to calculate $$\int_1^2 \frac{\phi(x) dx}{y(x)x^2\sqrt{x^2+(\frac{\phi(x)}{y(x)})^2}}$$
I tried $$y\phi'-y'\phi=y^2-9$$
On dividing by $y^2$
we get $$d(\frac{\phi}{y})=dx -\frac{9dx}{y^2}$$.The first two terms are integrable but I cannot understand how to deal with $\frac{9dx}{y^2}$.
I even tried differentiating the given equation to get$$y''\phi+y'\phi'-y'\phi'-y\phi''+2yy'=0$$.After cancelling we can rearrange to get$$\frac{d(\frac{\phi}{y})}{dx}=\frac{2dy}{y}.$$Unaware of how to handle such equation.
Please help.Thanks.
 A: $$y'(x)\phi(x)-y(x)\phi'(x)+y^2(x)=9 \tag 1$$
Prakhar Mishra gives an important additional  important information in comment : Both functions $y(x)$ and $\phi(x)$ are unknown. 
Consider an arbitrary function $f(x)$ and let $\quad y(x)=1+f(x)-f(1)$. 
Equation $(1)$ is a linear ODE easy to solve for $\phi(x)$ : 
$$\phi(x)=x\,y(x)-9\,y(x)\int\frac{dx}{(y(x))^2} \tag 2$$
Now, the function $\phi(x)$ is known. We can put it into $(3)$ and compute the integral:
$$\int_1^2 \frac{\phi(x) dx}{y(x)x^2\sqrt{x^2+(\frac{\phi(x)}{y(x)})^2}} \tag 3$$
With this method, we obtain $y(2)=1+f(2)-f(1)$ and the value of integral $(3)$, which is a solution of the problem. Since $f(x)$ is any function, the problem has an infinity of solutions.
For example, with $f(x)=x$ :
$y(x)=1+f(x)-f(1)=1+x-1=x$
$y(1)=1$ satisfies the first condition.
$y(2)=2$ is the answer yo the first part of the question.
$\phi(x)=x^2-9\,x\int\frac{dx}{x^2}=x^2+9+cx\quad$
We can take any one of these functions, for example $\phi(x)=x^2+9$.
The answer to the second part of the question is :
$$\int_1^2 \frac{(x^2-9) dx}{x^3\sqrt{x^2+\left(\frac{x^2-9}{x}\right)^2}} \simeq -0.474317$$
One can proceed on the same manner with other functions $f(x)$ arbitrary chosen and obtain as many solutions of the problem as we want.
