the question is as above. I used multiply both sides by the integrating factor I(x) and integrate both sides, but this method doesn't work, is there any way to solve this problem?
$x\ln(x)y'+y = xe^x$ solve $y$
0
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calculus
ordinary-differential-equations
1 Answers
3
If $x>0$ just multiply both sides by $x^{-1}$ $$\ln(x)y'+\frac1xy=e^x$$ Which is equivalent to $$\frac{d}{dx}[y\ln(x)]=e^x$$ Then, integration give us $$y\ln(x)=e^x+C,\qquad x>0$$