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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
MCQ
If $\int\limits_a^x {t\,y(t)dt} = x^2 + y (x)$ then $y$ as a function of $x$ is
  • ✓
    $y = 2 - (2 + a^2) {e^{\frac{{{x^2} - {a^2}}}{2}}}$
  • B
    $y = 1 - (2 + a^2)  {e^{\frac{{{x^2} - {a^2}}}{2}}}$
  • C
    $y = 2 - (1 + a^2)  {e^{\frac{{{x^2} - {a^2}}}{2}}}$
  • D
    none
✓

Answer

Correct option: A.
$y = 2 - (2 + a^2) {e^{\frac{{{x^2} - {a^2}}}{2}}}$
a
diff. both sides
$x y (x) = 2x - y' (x)$ 
hence $\frac{{dy}}{{dx}} - xy = - 2x$ 
$I.F = {e^{\int { - x\,dx} }} = {e^{\frac{{ - \;{x^2}}}{2}}}$ 
$y\,{e^{\frac{{ - \;{x^2}}}{2}}} = \int { - 2x\,{e^{\frac{{ - \;{x^2}}}{2}}}\,dx} $ 
$y\,{e^{\frac{{ - \;{x^2}}}{2}}} =  - 2x\,{e^{\frac{{ - \;{x^2}}}{2}}} + c$ 
$y = 2 + c\,{e^{\frac{{ - \;{x^2}}}{2}}}$ 
if $x = a$ 
$\Rightarrow a^2 + y = 0 \,\, \Rightarrow \,\, y = - a^2$ 
hence  $- a^2 = 2 + c\,{e^{\frac{{ - \;{a^2}}}{2}}}$ 
$c\,{e^{\frac{{ - \;{a^2}}}{2}}}= - (2 + a^2)$ 
$c = - (2 + a^2) {e^{\frac{{ - \;{a^2}}}{2}}}$ 
$y = 2 - (2 + a^2) {e^{\frac{{{x^2} - {a^2}}}{2}}}$

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