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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you post again your question? please be more clear.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=(\ln(x) + c)/x\] solution of: \[x^2y' + xy = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0prove it; \[y=(\ln(x)+c)/x\] \[xy=\ln(x)+c\] implicit derivative \[xy'+y=1/x\] multiply for x to both sides \[x^2y′+xy=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could you became my fan!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2y' + xy = 1\]\[x^2 \cdot \frac{c + \log{x}  1}{x^2} + x\frac{\log{x} + c}{x} = 1\]\[\log{x} + c  (c+\log{x}1) = 1\]\[1=1\] Thus \[\frac{\log{x}+c}{x}\] is a general solution of that differential equation.
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