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Question Number 80508 by M±th+et£s last updated on 03/Feb/20
solve the D.E   x^2 +(y^2 +1)dx+y(√(x^3 +1)) dy=0
$${solve}\:{the}\:{D}.{E}\: \\ $$$${x}^{\mathrm{2}} +\left({y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{y}\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dy}=\mathrm{0} \\ $$
Commented by mr W last updated on 03/Feb/20
syntax of equation is wrong.
$${syntax}\:{of}\:{equation}\:{is}\:{wrong}. \\ $$
Commented by M±th+et£s last updated on 03/Feb/20
so sorry but my English isn^, t good   and i am trying to be better
$${so}\:{sorry}\:{but}\:{my}\:{English}\:{isn}^{,} {t}\:{good}\: \\ $$$${and}\:{i}\:{am}\:{trying}\:{to}\:{be}\:{better} \\ $$
Commented by mr W last updated on 03/Feb/20
your english is no problem sir!  but the form of the equation   x^2 +(y^2 +1)dx+y(√(x^3 +1)) dy=0  makes no sense.  either you have:  x^2 +(y^2 +1)+y(√(x^3 +1)) (dy/dx)=0  or you have:  [x^2 +(y^2 +1)]dx+y(√(x^3 +1)) dy=0    please recheck!
$${your}\:{english}\:{is}\:{no}\:{problem}\:{sir}! \\ $$$${but}\:{the}\:{form}\:{of}\:{the}\:{equation}\: \\ $$$${x}^{\mathrm{2}} +\left({y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{y}\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dy}=\mathrm{0} \\ $$$${makes}\:{no}\:{sense}. \\ $$$${either}\:{you}\:{have}: \\ $$$${x}^{\mathrm{2}} +\left({y}^{\mathrm{2}} +\mathrm{1}\right)+{y}\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${or}\:{you}\:{have}: \\ $$$$\left[{x}^{\mathrm{2}} +\left({y}^{\mathrm{2}} +\mathrm{1}\right)\right]{dx}+{y}\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dy}=\mathrm{0} \\ $$$$ \\ $$$${please}\:{recheck}! \\ $$
Commented by M±th+et£s last updated on 04/Feb/20
you are right sir its my fault i forgat  the parentheses
$${you}\:{are}\:{right}\:{sir}\:{its}\:{my}\:{fault}\:{i}\:{forgat} \\ $$$${the}\:{parentheses} \\ $$

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