Question

solve using integrating factor y'=x-2y
tnx:)

Answer 1

y'+2y=x
y'+P(x)y=Q(x)

I=\(e^{\int P(x)dx}\)
=\(e^{\int 2dx}\)
= \(e^{2x}\)

\[e^{2x}y'+2e^{2x}y=e^{2x}x\]\[\frac{d e^{2x} y}{dx} =e^{2x}x\]\[\int\frac{d e^{2x} y}{dx} =\int e^{2x}x dx\]\[e^{2x} y=\frac{1}{4}e^{2x}(2x-1)+C\]\[y=\frac{1}{4}(2x-1)+C\]