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9. differential equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths9. differential equations1 Mark
MCQ
The solution of the equation $(x + 2{y^3})\frac{{dy}}{{dx}} - y = 0$ is
  • A
    $y(1 - xy) = Ax$
  • ✓
    ${y^3} - x = Ay$
  • C
    $x(1 - xy) = Ay$
  • D
    $x(1 + xy) = Ay$Where $A$ is any arbitrary constant
✓

Answer

Correct option: B.
${y^3} - x = Ay$
b
(b) $(x + 2{y^3})\frac{{dy}}{{dx}} = y$ ==> $\frac{{dy}}{{dx}} = \frac{y}{{x + 2{y^3}}}$

==> $\frac{{dx}}{{dy}} = \frac{{x + 2{y^3}}}{y}$ or $\frac{{dx}}{{dy}} - \frac{x}{y} = 2{y^2}$,

which is a linear equation of the form $\frac{{dx}}{{dy}} + Px = Q$

So, integrating factor $(I.F.)$$ = {e^{ - \int_{}^{} {\frac{1}{y}dy} }}$and solution is

$x\frac{1}{y} = \int_{}^{} {\frac{1}{y}2{y^2}dy + A = {y^2} + A} $ ==> $x = {y^3} + Ay$

==> ${y^3} - x = Ay;$where $A$ can be $ - ve$or $ + ve$.

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