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STD 12 - 8. Application and integration — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 8. Application and integration4 Marks
MCQ
A curve satisfying the initial condition $y(1)= 0$ satisfies the differential equation $x \frac{dy}{dx}= y -x^2$ the area bounded by the curve and the $x$ -axis is
  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{4}$
  • ✓
    $\frac{1}{6}$
✓

Answer

Correct option: D.
$\frac{1}{6}$
d
$x \frac{ dy }{ dx }= y - x ^2$

$\frac{ dy }{ dx }=\frac{ y }{ x }- x$

$\frac{ y }{ x }= m$

$\frac{ dy }{ dx }= x \frac{ dm }{ dx }+ m$

$x \frac{ dm }{ dx }+ m = m - x$

$=\frac{ dm }{ dx }=-1$

$m =- x + c$

$y =- x ^2+ cx$

$y (1)=0$

$0=-1+ c$

$c =1$

$\int \limits_0^1\left(- x ^2+ x \right) dx$

$=-\frac{1}{3}+\frac{1}{2}$

$y =- x ^2+ x$

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