xy"+2y'=0 reduced into y
Let us solve xy′′+2y′=0.xy''+2y'=0.xy′′+2y′=0. Let z=y′.z=y'.z=y′. Then xz′+2z=0.xz'+2z=0.xz′+2z=0. It follows that xdzdx=−2z,x\frac{dz}{dx}=-2z,xdxdz=−2z, and hence dzz=−2dxx.\frac{dz}{z}=-2\frac{dx}{x}.zdz=−2xdx. We have that ∫dzz=−2∫dxx,\int\frac{dz}{z}=-2\int\frac{dx}{x},∫zdz=−2∫xdx, and thus ln∣z∣=−2ln∣x∣+ln∣C∣=ln∣Cx2∣\ln|z|=-2\ln|x|+\ln|C|=\ln|\frac{C}{x^2}|ln∣z∣=−2ln∣x∣+ln∣C∣=ln∣x2C∣ . We conclude that z=Cx2.z=\frac{C}{x^2}.z=x2C. Then y′=Cx2,y'=\frac{C}{x^2},y′=x2C, and hence the solution is y=−Cx+C2.y=-\frac{C}{x}+C_2.y=−xC+C2.
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