I.F. of x+1dydx-y=e2x(x+1)2

dydx−11+x.y= e2x(x+1)I.F=e∫−11+x dx=e−log(1+x)=elog (11+x)=11+x

I.F. of x+1dydx-y=e2x(x+1)2

dydx−11+x.y= e2x(x+1)I.F=e∫−11+x dx=e−log(1+x)=elog (11+x)=11+x

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I.F. of $(x + 1) \frac{dy}{dx} - y = e^{2 x} {(x + 1)}^{2}$

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–(x + 1)

log(x+1)

$\frac{e^{2 x}}{2}$

$\frac{1}{x + 1}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} - \frac{1}{1 + x} . y = e^{2 x} (x + 1) \\ I . F = e^{\int \frac{- 1}{1 + x} d x} \\ = e^{- \log (1 + x)} \\ = e^{\log (\frac{1}{1 + x})} \\ = \frac{1}{1 + x} \end{array}$

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