Solution of differential equation d y/d x=e^{x-y}(e^x-e^y)

Correct Answer is : eyeex=exeex−eex+ce^y e^{e^x} = e^x e^{e^x} - e^{e^x} + ceyeex=exeex−eex+c

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

e^{x-y} = e^{e^x} + x² e^{e^x} + c

Correct Answer is : eyeex=exeex−eex+ce^y e^{e^x} = e^x e^{e^x} - e^{e^x} + ceyeex=exeex−eex+c

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CGPET 2022 Solved Paper

Section: Mathematics

Question : 115 of 150

Marks: +1, -0

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

Solution:

We have,

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

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

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

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

Put

e^y = v

\Rightarrow e^y \frac{d y}{d x} = \frac{d v}{d x}

\therefore \frac{d v}{d x} + e^x v = e^{2x}

Now, IF

=e^{\int e^x dx}=e^{e^x}

Hence, solution is given as

v \cdot e^{e^x} = \int e^{2x} \cdot e^{e^x} dx + C

= \int e^x e^{e^x} e^x dx + C

Put

e^x = t

\Rightarrow e^x dx = dt

\therefore v e^{e^x} = \int t e^t dt + c

v e^{e^x} = t e^t - \int e^t \cdot 1 dt + c

v e^{e^x} = t e^t - e^t + c

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

which is required solution.

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