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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
Solution of $ydx - xdy = {x^2}ydx$ is
  • A
    $y{e^{{x^2}}} = c{x^2}$
  • B
    $y{e^{ - {x^2}}} = c{x^2}$
  • ✓
    ${y^2}{e^{{x^2}}} = c{x^2}$
  • D
    ${y^2}{e^{ - {x^2}}} = c{x^2}$
✓

Answer

Correct option: C.
${y^2}{e^{{x^2}}} = c{x^2}$
c
(c) Given equation can be written as $\left( {\frac{{1 - {x^2}}}{x}} \right)dx = \frac{{dy}}{y}$

After integration, we get $\log x - \frac{{{x^2}}}{2} = \log y + \log c$

==> $\log {x^2} - \log {y^2} + \log c = {x^2}$ ==> $\log \frac{{c{x^2}}}{{{y^2}}} = {x^2}$

==> $\frac{{c{x^2}}}{{{y^2}}} = {e^x}^2$ ==> $c{x^2} = {y^2}{e^{{x^2}}}$.

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