Final answer to the problem
$-\ln\left|y\right|+\ln\left|y-1\right|=\ln\left|x\right|+C_0$
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Step-by-step Solution
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- Equation différentielle exacte
- Équation différentielle linéaire
- Équation différentielle séparable
- Equation différentielle homogène
- Produit de binômes avec terme commun
- Méthode FOIL
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1
Apply the formula: $a\frac{b}{c}$$=\frac{ba}{c}$, where $a=x$, $b=dy$ and $c=dx$
$\frac{xdy}{dx}+y=y^2$
Learn how to solve différenciation logarithmique problems step by step online.
$\frac{xdy}{dx}+y=y^2$
Learn how to solve différenciation logarithmique problems step by step online. xdy/dx+y=y^2. Apply the formula: a\frac{b}{c}=\frac{ba}{c}, where a=x, b=dy and c=dx. Apply the formula: x+a=b\to x=b-a, where a=y, b=y^2, x+a=b=\frac{xdy}{dx}+y=y^2, x=\frac{xdy}{dx} and x+a=\frac{xdy}{dx}+y. Group the terms of the differential equation. Move the terms of the y variable to the left side, and the terms of the x variable to the right side of the equality. Simplify the expression \frac{1}{y^2-y}dy.
Final answer to the problem
$-\ln\left|y\right|+\ln\left|y-1\right|=\ln\left|x\right|+C_0$
