Question

prove y=(ln(x) + c)/x is a soln. of: x^2y' + xy = 1

Answer 1

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Answer 2

\[y=(\ln(x) + c)/x\]
solution of:
\[x^2y' + xy = 1\]

Answer 3

prove it;
\[y=(\ln(x)+c)/x\]
\[xy=\ln(x)+c\]
implicit derivative
\[xy'+y=1/x\]
multiply for x to both sides
\[x^2y′+xy=1\]

Answer 4

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Answer 5

\[x^2y' + xy = 1\]\[x^2 \cdot -\frac{c + \log{x} - 1}{x^2} + x\frac{\log{x} + c}{x} = 1\]\[\log{x} + c - (c+\log{x}-1) = 1\]\[1=1\]
Thus \[\frac{\log{x}+c}{x}\] is a general solution of that differential equation.