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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
For $x \in R,x \ne 0$, if $y(x)$ is a differentiable function such that $x\int\limits_1^x {y\left( t \right)} dt = \left( {x + 1} \right)\int\limits_1^x {ty\left( t \right)} dt$ , then $y(x)$ equals (where $C$ is a constant)
  • A
    $c{x^3}{e^{\frac{1}{x}}}$
  • B
    $\frac{c}{{{x^2}}}{e^{ - \frac{1}{x}}}$
  • C
    $\frac{c}{{{x}}}{e^{ - \frac{1}{x}}}$
  • ✓
    $\frac{{c{e^{ - \frac{1}{x}}}}}{{{x^3}}}$
✓

Answer

Correct option: D.
$\frac{{c{e^{ - \frac{1}{x}}}}}{{{x^3}}}$
d
$x\int\limits_1^x {y(t)dt = x\int\limits_1^x {ty(t)dt + \int\limits_1^x {ty(t)dt} } } $

Differentiate $w r:$ to $x$

$\int\limits_1^x {y(t)dt + x[y(x) - y(1)]} $

$ = \int\limits_1^x {ty(t)dt + x[xy(x) - y(1)] + xy(x) - y(1)} $

$\int\limits_1^x {y(t)dt}  = \int\limits_1^x t y(t)dt + {x^2}y(x) - y(1)$

Diff. again wr. to$x $ $y\left( x \right) - y\left( 1 \right) = xy\left( x \right) - y\left( 1 \right) + 2xy\left( x \right) + {x^2}{y^1}$

$(x)$

$(1-3 x) y(x)=x^{2} y^{1}(x)$

$\frac{y^{1}(x)}{y(x)}=\frac{1-3 x}{x^{2}}$

$\frac{{1dy}}{{ydx}} = \frac{{1 - 3x}}{{{x^2}}}$

$ \Rightarrow \ln y =  - \frac{1}{x} - 3\ln x$

$\ln \left(y x^{3}\right)=-\frac{1}{x}$

$y x^{3}=-e^{-1 / x}$

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