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The given differential equation is $(1+x^{2})\frac{dy}{dx}+2xy-4x^{2}=0$. We can rewrite it as: $$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2}$$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$. The integrating factor (IF) is given by $e^{\int P(x) dx}$. $$\int P(x) dx = \int \frac{2x}{1+x^2} dx$$ Let $u = 1+x^2$, then $du = 2x dx$. $$\int \frac{du}{u} = \ln|u| = \ln(1+x^2)$$ Therefore, the integrating factor is: $$IF = e^{\ln(1+x^2)} = 1+x^2$$
Multiplying the differential equation by the integrating factor $(1+x^2)$, we get: $$(1+x^2)\frac{dy}{dx} + 2xy = 4x^2$$ $$\frac{d}{dx}[y(1+x^2)] = 4x^2$$
Integrating both sides with respect to $x$, we have: $$\int \frac{d}{dx}[y(1+x^2)] dx = \int 4x^2 dx$$ $$y(1+x^2) = \frac{4}{3}x^3 + C$$
Given the initial condition $y(0) = 0$, we substitute $x=0$ and $y=0$ into the general solution: $$0(1+0^2) = \frac{4}{3}(0)^3 + C$$ $$0 = 0 + C$$ $$C = 0$$
Substituting $C=0$ into the general solution, we get the particular solution: $$y(1+x^2) = \frac{4}{3}x^3$$ $$y = \frac{4x^3}{3(1+x^2)}$$
Final Answer: $y = \frac{4x^3}{3(1+x^2)}$
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