y^(')=-(2y)/(x) Cpx^(p-1)=-2Cx^(p) pCx^(p-1)=-2Cx^(p-1) pCx^(p-1)=-2Cx^(p)Could you explain how the x vanishes in the denominator and why does the left hand side have predict at the start but the right does not?

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Solution for - y^(')=-(2y)/(x) Cpx^(p-1)=-2Cx^(p) pCx^(p-1)=-2Cx^(p-1) pCx^(p-1)=-2Cx^(p)Could you explain how the

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This an additional answer to - y^(')=-(2y)/(x) Cpx^(p-1)=-2Cx^(p) pCx^(p-1)=-2Cx^(p-1) pCx^(p-1)=-2Cx^(p)Could you explain how the

(Solved): y^(')=-(2y)/(x) Cpx^(p-1)=-2Cx^(p) pCx^(p-1)=-2Cx^(p-1) pCx^(p-1)=-2Cx^(p)Could you explain how the ...

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