If- ey 1 - x - 1- then show that d2ydx2 - dydx 2

Given: e^y (x + 1) = 1 Taking logarithm of both sides, we get y log #160;e + log (x + 1) = log #160;1 ⇒ y + log (x + 1) = 0 ⇒ y = log ( ...

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Standard XII

Mathematics

General Solution of a Differential Equation

If, ey 1 + ...

Question

If, $e^{y} (1 + x) = 1$ , then show that
$\frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}$

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Solution

Given: $e^{y} (x + 1) = 1$
Taking logarithm of both sides, we get
$y log e + log (x + 1) = log 1$
$\Rightarrow y + log (x + 1) = 0$
$\Rightarrow y = - log (x + 1)$
Differentiating w.r.t. $x$ , we get,
$\frac{d y}{d x} = - \frac{1}{x + 1}$ ............ (i)
Diffeerentiating (i) again w.r.t. $x$ , we get,
$\frac{d^{2} y}{d x^{2}} = \frac{1}{(x + 1)^{2}} = {(\frac{- 1}{x + 1})}^{2}$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}$ [From (i)]

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If, ey(1+x)=1, then show thatd2ydx2=(dydx)2

Given: ey(x+1)=1Taking logarithm of both sides, we getyloge+log(x+1)=log1⇒y+log(x+1)=0⇒y=−log(x+1)Differentiating w.r.t. x, we get,dydx=−1x+1 ............ (i)Diffeerentiating (i) again w.r.t. x, we get,d2ydx2=1(x+1)2=(−1x+1)2⇒d2ydx2=(dydx)2 [From (i)]

If, $e^{y} (1 + x) = 1$ , then show that
$\frac{d^{2} y}{d x^{2}} = {(\frac{d y}{d x})}^{2}$