Answer:
Step-by-step explanation:
[tex]\frac{dy}{dx} =\frac{x^2+1}{e^y} \\separating~the~variables~and~integrating\\\int\ {e^y} \, dy=\int\ {(x^2+1)} \, dx \\e^y=\frac{x^3}{3} +x+c\\x=1,y=0\\e^0=\frac{1}{3} +1+c\\c=-\frac{1}{3} \\e^y=\frac{x^3}{3} +x-\frac{1}{3} \\when x=2\\e^y=\frac{2^3}{3} +2-\frac{1}{3} =\frac{8+6-1}{3} =\frac{13}{3} \\y=log(\frac{13}{3} )[/tex]
