xy' - (x+2)y = 0, x0 = 0
xy′−(x+2)y=0xy' - (x+2)y = 0xy′−(x+2)y=0
separate variables : dyy=x+2xdx⇒d(ln(y))=(1+2x)dx\frac{dy}{y} = \frac{x+2}{x} dx \Rightarrow d(ln(y)) = (1 + \frac{2}{x})dxydy=xx+2dx⇒d(ln(y))=(1+x2)dx
integrate : ln(y)=x+2⋅ln(x)+constln(y) = x + 2\cdot ln(x) + constln(y)=x+2⋅ln(x)+const , y=x2⋅ex⋅Cy = x^2 \cdot e^{x} \cdot Cy=x2⋅ex⋅C , where C = const
ANSWER : y=x2⋅ex⋅Cy = x^2 \cdot e^{x} \cdot Cy=x2⋅ex⋅C , where C = const
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