Question #211330

xy"+2y'=0 reduced into y


1
Expert's answer
2021-07-04T17:27:25-0400

Let us solve xy+2y=0.xy''+2y'=0. Let z=y.z=y'. Then xz+2z=0.xz'+2z=0. It follows that xdzdx=2z,x\frac{dz}{dx}=-2z, and hence dzz=2dxx.\frac{dz}{z}=-2\frac{dx}{x}. We have that dzz=2dxx,\int\frac{dz}{z}=-2\int\frac{dx}{x}, and thus lnz=2lnx+lnC=lnCx2\ln|z|=-2\ln|x|+\ln|C|=\ln|\frac{C}{x^2}| . We conclude that z=Cx2.z=\frac{C}{x^2}. Then y=Cx2,y'=\frac{C}{x^2}, and hence the solution is y=Cx+C2.y=-\frac{C}{x}+C_2.



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