Question #178309

xy' - (x+2)y = 0, x0 = 0


1
Expert's answer
2021-05-07T09:03:38-0400

xy(x+2)y=0xy' - (x+2)y = 0

separate variables : dyy=x+2xdxd(ln(y))=(1+2x)dx\frac{dy}{y} = \frac{x+2}{x} dx \Rightarrow d(ln(y)) = (1 + \frac{2}{x})dx

integrate : ln(y)=x+2ln(x)+constln(y) = x + 2\cdot ln(x) + const , y=x2exCy = x^2 \cdot e^{x} \cdot C , where C = const


ANSWER : y=x2exCy = x^2 \cdot e^{x} \cdot C , where C = const


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