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Tess tried to solve the differential equation $\frac{d y}{d x}=2 x y-6 x$. This is her work:$\newline$$$\frac{d y}{d x}=2 x y-6 x$$$\newline$Step $1$: $\quad \frac{d y}{d x}=(y-3) 2 x$$\newline$Step $2$: $\quad \int \frac{1}{y-3} d y=\int 2 x d x$$\newline$Step $3$: $\quad \ln |y-3|=x^{2}+C_{1}$$\newline$Step $4$: $\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}$$\newline$Step $5$: $\quad|y-3|=e^{x^{2}}+C_{1}$$\newline$Step $6$: $\quad y-3= \pm e^{x^{2}}+C_{2}$$\newline$Step $7$: $\quad y= \pm e^{x^{2}}+C$$\newline$Is Tess's work correct? If not, what is her mistake?$\newline$Choose $1$ answer:$\newline$(A) Tess's work is correct.$\newline$(B) Step $1$ is incorrect. We're not allowed to factor an expression when solving differential equations.$\newline$(C) Step $4$ is incorrect. The right-hand side of the equation should be $e^{x^{2}+C_{1}}$.$\newline$(D) Step $7$ is incorrect. Tess didn't account for adding $3$ to the right side.