enay800 2 years ago dy/dx ln(xy)=xy^2-x

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1. Azteck

What are you meant to find or solve?

2. Azteck

$\large \frac{dy}{dx} ln(xy)=xy^2-x$ $\large \frac{dy}{dx}[lnx+lny]=xy^2-x$ $\large \frac{1}{x}=xy^2-x$ $\large 1=x^2y^2-x^2$ $\large y^2=\frac{x^2+1}{x^2}$ $\large y=\pm \frac{\sqrt{x^2+1}}{x}$

3. abdullatif92211

this question can be solved using bernaoulli method