anonymous
  • anonymous
prove y=(ln(x) + c)/x is a soln. of: x^2y' + xy = 1
Mathematics
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anonymous
  • anonymous
prove y=(ln(x) + c)/x is a soln. of: x^2y' + xy = 1
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
Could you post again your question? please be more clear.
anonymous
  • anonymous
\[y=(\ln(x) + c)/x\] solution of: \[x^2y' + xy = 1\]
anonymous
  • anonymous
prove it; \[y=(\ln(x)+c)/x\] \[xy=\ln(x)+c\] implicit derivative \[xy'+y=1/x\] multiply for x to both sides \[x^2y′+xy=1\]

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anonymous
  • anonymous
could you became my fan!
apples
  • apples
\[x^2y' + xy = 1\]\[x^2 \cdot -\frac{c + \log{x} - 1}{x^2} + x\frac{\log{x} + c}{x} = 1\]\[\log{x} + c - (c+\log{x}-1) = 1\]\[1=1\] Thus \[\frac{\log{x}+c}{x}\] is a general solution of that differential equation.

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