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Popular Calculus >

y^'=-((2x+ln(y))y)/x

Solution

y^{^{'}} = - \frac{( 2 x + ln ( y ) ) y}{x}

Solution

y = e^{\frac{c _{1}}{x} - x}

Solution steps

y^{'} (x) = - \frac{( 2 x + ln ( y ) ) y}{x}

Solve exact ODE:

y = e^{\frac{c _{1}}{x} - x}

y = e^{\frac{c _{1}}{x} - x}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'=-((2x+ln(y))y)/x ?
The solution for y^'=-((2x+ln(y))y)/x is y=e^{(c_{1})/x-x}