The integrating factor of the differential equation d y-d x1-x-x y-1-x isA- 1-x exB- x-1 exC- 1- x e - xD- 1- x e - x

The correct option is C (1+x)e^ xdydx(1+x) xy=1 x ⇒dydx x1+xy=1 x1+x Hence, I.F. =exp nbsp; (∫ x1+x dx ) =exp( ∫(1 1/1+x)dx) =e^ (x ln (1+x))=e^ x(1+x) ...

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Standard XII

Mathematics

Integrating Factor

The integrati...

Question

The integrating factor of the differential equation $\frac{d y}{d x} (1 + x) - x y = 1 - x$ is

(1 + x) e^{x}

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(x - 1) e^{x}

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(1 + x) e^{- x}

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(1 - x) e^{- x}

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Solution

The correct option is C $(1 + x) e^{- x}$
$\frac{d y}{d x} (1 + x) - x y = 1 - x$
$\Rightarrow \frac{d y}{d x} - \frac{x}{1 + x} y = \frac{1 - x}{1 + x}$

Hence, I.F. $= exp (\int \frac{- x}{1 + x} d x)$

$= exp (- \int (1 - \frac{1}{1 + x}) d x)$

$= e^{- (x - ln (1 + x))} = e^{- x} (1 + x)$

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The integrating factor of the differential equation dydx(1+x)−xy=1−x is

The correct option is C (1+x)e−xdydx(1+x)−xy=1−x ⇒dydx−x1+xy=1−x1+x Hence, I.F. =exp(∫−x1+xdx) =exp(−∫(1−11+x)dx) =e−(x−ln(1+x))=e−x(1+x)

The integrating factor of the differential equation $\frac{d y}{d x} (1 + x) - x y = 1 - x$ is

The correct option is C $(1 + x) e^{- x}$
$\frac{d y}{d x} (1 + x) - x y = 1 - x$
$\Rightarrow \frac{d y}{d x} - \frac{x}{1 + x} y = \frac{1 - x}{1 + x}$

Hence, I.F. $= exp (\int \frac{- x}{1 + x} d x)$

$= exp (- \int (1 - \frac{1}{1 + x}) d x)$

$= e^{- (x - ln (1 + x))} = e^{- x} (1 + x)$