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bernoulli (dy)/(dx)=(y^2)/(x^3),y(1)=1

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∫ 2x³−2x²+1x²−x dx

Solution

x²−ln|x|+ln|x−1|+C

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∫ 2x³−2x²+1x²−x dx

Prendre la fraction partielle de 2x³−2x²+1x²−x : 2x−1x +1x−1

=∫ 2x−1x +1x−1 dx

Appliquer la règle de la somme: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ 2xdx−∫ 1x dx+∫ 1x−1 dx

∫ 2xdx=x²

∫ 1x dx=ln|x|

∫ 1x−1 dx=ln|x−1|

=x²−ln|x|+ln|x−1|

Ajouter une constante à la solution

=x²−ln|x|+ln|x−1|+C

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

bernoulli (dy)/(dx)=(y^2)/(x^3),y(1)=1

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