<s>[INST] Stuck on this linear equation $y'=(3x^2-e^x)/2y-5$ Given this separable equation: $y'=(3x^2-e^x)/2y-5$.
I did integration, I found my constant $c=-3$.
And now I am stuck on this part: $y^2-5y=x^3-e^x-3$.
No matter what I do, I cannot come to this correct answer, according to the textbook, $y=-5/2-sqrt(x^3-e^x+13/4)$.
 [/INST] You did it right. Just see that : $$y^2-5y+\frac{25}{4}=x^3-e^x-3+\frac{25}{4}$$ so $$\left(y+\frac{5}{2}\right)^2=x^3-e^x-3+\frac{25}{4}=x^3-e^x+\frac{13}{4}$$ so one solution may arise of a form: $$y=-\frac{5}{2}+\sqrt{x^3-e^x+\frac{13}{4}}$$
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