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Variation of parametersFirst, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation givesy′=2xy⇔dydx=2xy⇔dyy=2xdxNow, the variables are separated, x appears only on the right side, and y only on the left.Integrate the left side in relation to y, and the right side in relation to x∫dyy=2∫xdxWhich isln|y|=x2+cBy taking exponents, we obtain|y|=ex2+c=ex2⋅ecHence,we obtainy=Cex2where C=±ecandyc=ex2 is the complementary solution .Next, we need to find the particular solution ypTherefore, we consider uyc, and try to find u, a function of x, that will make this work.Let's assume that uyc is a solution of the given equation. Hence, it satisfies the given equation.Substituting uyc, and its derivative in the equation gives(uyc)′+2xuyc=ex2u′yc+uyc′+2xuyc=ex2u′yc+u(yc′+2xyc)⏟=0 =ex2since yc is a solutionTherefore,u′yc=ex2⇒u′=ex2ycwhich givesu=∫ex2ycdxNow, we can find the function u:u=∫(ex2ex2)dx=∫dx=x+cSince we need to find only one function that will male this work, we don’t need to introduce the constant of integration c. Hence,u=xRecall that yp=uycThereforeyp=xex2The general solution isy=Cyc+yp=Cex2+xex2=ex2(x+C)Integration Factor techniqueThis equation is linear with P(x)=−2xandQ(x)=ex2Hence,h=−
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