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

1. anonymous

2. anonymous

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

3. anonymous

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

4. anonymous

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5. 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.