I have $\frac{dy}{dx}=5xy + \sin x$, $y(0) = 1$. I am asked to use an integrating factor. What I did:
$\frac{dy}{dx}-5xy = \sin x \\ \text{Integrating factor:} \ e^{\int{-5x\ dx}} = e^{-\frac{5}{2}x^2} \\ \frac{d}{dx}\left[e^{-\frac{5}{2}x^2}y\right] = e^{-\frac{5}{2}x^2}\sin x \\ e^{-\frac{5}{2}x^2}y = \int e^{-\frac{5}{2}x^2}\sin x \ dx \\ y = e^{\frac{5}{2}x^2}\int e^{-\frac{5}{2}x^2}\sin x \ dx$
However, in my previous thread, How to solve $\frac{dy}{dx}=5xy + \sin x$?, few users said $y = e^{\frac{5}{2}x^2}\big(1+\int_0^x e^{-\frac{5}{2}t^2}\sin t \, dt\big)$ or $y = e^{\frac{5}{2} x^2} (\int e^{-\frac{5}{2} x^2} \sin x\ dx + C)$
But Im wondering, where does the $1$ or $c$ come from? Shouldnt it just be $y = e^{\frac{5}{2}x^2}\int e^{-\frac{5}{2}x^2}\sin x \ dx$?
Edit: $y = e^{\frac{5}{2}x^2}\int e^{-\frac{5}{2}x^2}\sin x \ dx + Ce^{\frac{5}{2}x^2}\\ \text{using } y(0) =1 \\c=1\\ y = e^{\frac{5}{2}x^2}\left(\int e^{-\frac{5}{2}x^2}\sin x \ dx + 1\right)$