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Theorems for Differentiability

Solution of t...

Question

Solution of the differential equation $(\frac{d y}{d x}) = e^{x - y} (e^{x} - e^{y})$ is:
(where $c$ is integration constant)

A

e^{y} e^{e^{x}} = e^{x} e^{e^{x}} + c

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B

e^{y} e^{e^{x}} = e^{e^{x}} + c

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C

e^{y} e^{x} = e^{x} e^{e^{x}} - e^{e^{x}} + c

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D

e^{y} e^{e^{x}} = e^{x} e^{e^{x}} - e^{e^{x}} + c

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Solution

The correct option is D $e^{y} e^{e^{x}} = e^{x} e^{e^{x}} - e^{e^{x}} + c$
Multiplying the given equation by $e^{y}$ , we get
$e^{y} \frac{d y}{d x} + e^{x} e^{y} = e^{2 x} \dots (1)$
Putting $e^{y} = v,$ so that $e^{y} \frac{d y}{d x} = \frac{d v}{d x},$
and equation $(1)$ transform to $\frac{d v}{d x} + e^{x} v = e^{2 x}$
On comparing with $\frac{d v}{d x} + P v = Q,$
We get $P = e^{x}$ and $Q = e^{2 x}$
$I . F . = e^{\int e^{x} d x} = e^{e^{x}}$

Hence solution is $v e^{e^{x}} = \int e^{2 x} e^{e^{x}} d x$
$\Rightarrow v e^{e^{x}} = \int (e^{x}) e^{(e^{x})} (e^{x} d x)$
Let $e^{x} = t \Rightarrow e^{x} d x = d t$
$\Rightarrow v e^{e^{x}} = \int t e^{t} d t$
$\Rightarrow e^{y} e^{e^{x}} = t e^{t} - e^{t} + c$
$\Rightarrow e^{y} e^{e^{x}} = e^{x} e^{e^{x}} - e^{e^{x}} + c$

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Theorems for Differentiability

Standard XII Mathematics

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