Hello Thingsto Do,
I would first write the ODE in the form $$\frac{dy}{dx}=f(x,y)$$:
$$\frac{dy}{dx}=\frac{104xy-y^2}{x^2}=104\frac{y}{x}-\left(\frac{y}{x} \right)^2$$
Using the substitution $$u=\frac{y}{x}\,\therefore\,\frac{dy}{dx}=u+x\frac{du}{dx}$$ we may write:
$$u+x\frac{du}{dx}=104u-u^2$$
$$x\frac{du}{dx}=103u-u^2$$
Separating variables, we have:
$$\frac{1}{103u-u^2}\,du=\frac{1}{x}\,dx$$
Using the Heaviside cover-up method on the left side, we may obtain the partial fraction decomposition. Factoring the denominator, we may then assume it takes the following form:
$$\frac{1}{u(103-u)}=\frac{A}{u}+\frac{B}{103-u}$$
Covering up the factor $u$ in the denominator of the left side, and evaluating what is left for $u=0$ we obtain:
$$A=\frac{1}{103}$$
Covering up the factor $103-u$ in the denominator of the left side, and evaluating what is left for $u=103$ we obtain:
$$B=\frac{1}{103}$$
Hence:
$$\frac{1}{u(103-u)}=\frac{1}{103}\left(\frac{1}{u}+\frac{1}{103-u} \right)$$
And so the ODE becomes:
$$\frac{1}{103}\left(\frac{1}{u}-\frac{1}{u-103} \right)\,du=\frac{1}{x}\,dx$$
Integrating, we have:
$$\frac{1}{103}\int\left(\frac{1}{u}-\frac{1}{u-103} \right)\,du=\int\frac{1}{x}\,dx$$
$$\ln\left|\frac{u}{u-103} \right|=103\ln\left|Cx \right|=\ln\left|Cx^{103} \right|$$
$$\frac{u}{u-103}=Cx^{103}$$
Solve for $u$:
$$u=\frac{103Cx^{103}}{Cx^{103}-1}$$
Back-substitute for $u$:
$$\frac{y}{x}=\frac{103Cx^{103}}{Cx^{103}-1}$$
Hence:
$$y(x)=\frac{103Cx^{104}}{Cx^{103}-1}$$
Use the initial values to determine the parameter $C$:
$$y(1)=\frac{103C}{C-1}=2\,\therefore\,C=-\frac{2}{101}$$
Thus, the solution satisfying the IVP is:
$$y(x)=\frac{103\left(-\frac{2}{101} \right)x^{104}}{\left(-\frac{2}{101} \right)x^{103}-1}=\frac{206x^{104}}{2x^{103}+101}$$
