The correct option is B e^y = c. exp ( e^x) + e^x 1 dydx = e^x y (e^x e^y) dy dx = e^ x e^ y (e^ x e^ y ) e^ y dy dx = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Composite Function

dy/dx = ex-y ...

Question

$\frac{d y}{d x} = e^{x - y} (e^{x} - e^{y})$

e^{x} = c . e x p (e^{y}) + e^{y} + 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{x} = c . e x p (- e^{y}) + e^{y} - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{y} = c . e x p (- e^{x}) + e^{x} - 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

e^{y} = c . e x p (e^{x}) + e^{x} + 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $e^{y} = c . e x p (- e^{x}) + e^{x} - 1$
$\frac{d y}{d x} = e^{x - y} (e^{x} - e^{y})$

$\frac{d y}{d x} = \frac{e^{x}}{e^{y}} (e^{x} - e^{y})$

$e^{y} \frac{d y}{d x} = e^{2 x} - e^{x} e^{y}$

$e^{y} \frac{d y}{d x} + e^{x} e^{y} = e^{2 x}$

Put $e^{y} = v$
$e^{y} \frac{d y}{d x} = \frac{d v}{d x}$

$e^{y} \frac{d v}{d x} + {v e}^{x} = e^{2 x}$
which is a linear differential eqn with $v$ as dependent variable.
Here, $P = e^{x}; Q = e^{2 x}$

Integrating factor $I . F . = e^{\int} e^{x} d x = e^{e^{x}}$
So, the solution of given differential eqn is
$v . e^{e^{x}} = \int e^{e^{x}} e^{2 x} d x + C$

Put $e^{x} = t$ in the above integral
$\Rightarrow e^{x} d x = d t$

$v . e^{e^{x}} = \int e^{t} t d t + C$
$\Rightarrow e^{y} e^{e^{x}} = t e^{t} - e^{t} + C$
$\Rightarrow e^{y} e^{e^{x}} = e^{x} e^{e^{x}} - e^{e^{x}} + C$
$\Rightarrow e^{y} = e^{x} - 1 + C e^{- e^{x}}$

Suggest Corrections