Answer to Question #220816 in Calculus for Unknown346307

Question #220816

Given that y = x is a particular solution of the differential

equation y''-x2y'

+ xy = 0. Find its general solution.



1
Expert's answer
2021-07-28T13:43:45-0400

We make the substitution y=y1u=xuy=y_1u=xu

Then

y=u+xuy'=u+xu'

y=u+u+xu=2u+xuy''=u'+u'+xu''=2u'+xu''

Substitute


2u+xux2ux3u+x2u=02u'+xu''-x^2u-x^3u'+x^2u=0

xu+(2x3)u=0xu''+(2-x^3)u'=0

Let p=u.p=u'. Then u=p.u''=p'.


xp+(2x3)p=0xp'+(2-x^3)p=0

dpp=x32xdx\dfrac{dp}{p}=\dfrac{x^3-2}{x}dx

Integrate both sides


dpp=x32xdx\int\dfrac{dp}{p}=\int\dfrac{x^3-2}{x}dx

lnp=x332lnx+lnC\ln|p|=\dfrac{x^3}{3}-2\ln|x|+\ln C


p=c1ex3/3x2p=\dfrac{c_1e^{x^3/3}}{x^2}

u=c1ex3/3x2dxu=\int \dfrac{c_1e^{x^3/3}}{x^2}dx

y=xc1ex3/3x2dxy=x\int \dfrac{c_1e^{x^3/3}}{x^2}dx


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