Solve (সমাধান করো ): $$ \frac{d y}{d x}=e^{x+y-1} $$

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We are given the differential equation

$$
\frac{dy}{dx}=e^{x+y-1}.
$$

**Step 1: Separate the variables.**

Note that

$$
e^{x+y-1} = e^{x-1} \cdot e^y.
$$

Thus, the equation can be rewritten as

$$
\frac{dy}{dx}=e^{x-1}\cdot e^y.
$$

To separate the variables, divide both sides by $$ e^y $$ and multiply by $$ dx $$ to obtain

$$
\frac{dy}{e^y} = e^{x-1}\,dx.
$$

Since $$ \frac{1}{e^y}=e^{-y} $$, the equation becomes

$$
e^{-y}\,dy = e^{x-1}\,dx.
$$

**Step 2: Integrate both sides.**

Integrate the left side with respect to $$ y $$ and the right side with respect to $$ x $$:

$$
\int e^{-y}\,dy = \int e^{x-1}\,dx.
$$

**Step 3: Compute the integrals.**

For the left-hand side:

$$
\int e^{-y}\,dy = -e^{-y} + C_1,
$$

where $$ C_1 $$ is an integration constant.

For the right-hand side:

$$
\int e^{x-1}\,dx = e^{-1}\int e^x\,dx = e^{-1}\,e^x + C_2 = e^{x-1} + C_2,
$$

where $$ C_2 $$ is another integration constant.

**Step 4: Combine the results.**

We have

$$
-e^{-y} + C_1 = e^{x-1} + C_2.
$$

We can combine the arbitrary constants $$ C_1 $$ and $$ C_2 $$ into a single constant $$ C $$ by letting $$ C = C_2 - C_1 $$. Rearranging the terms, we have

$$
-e^{-y} = e^{x-1}+C.
$$

Multiplying every term by $$-1$$ yields

$$
e^{-y} = -e^{x-1} + \tilde{C},
$$

where $$ \tilde{C} = -C $$ is still an arbitrary constant. It is conventional to denote an arbitrary constant simply as $$ C $$. Therefore, we write

$$
e^{-y} = C - e^{x-1}.
$$

**Step 5: Solve for $$ y $$ (if desired).**

Taking the natural logarithm on both sides, we obtain

$$
-y = \ln\left(C - e^{x-1}\right).
$$

Thus, the solution for $$ y $$ is

$$
y = -\ln\left(C - e^{x-1}\right),
$$

with the implicit condition that $$ C - e^{x-1} > 0 $$.

**Final Answer:**

$$
\boxed{e^{-y}=C-e^{x-1} \quad \text{or equivalently} \quad y=-\ln\left(C-e^{x-1}\right).}
$$
The solution to the differential equation $$ \frac{dy}{dx} = e^{x+y-1} $$ is: $$ e^{-y} = C - e^{x-1} \quad \text{or equivalently} \quad y = -\ln\left(C - e^{x-1}\right), $$ where $$ C $$ is an arbitrary constant.

Solve (সমাধান করো ): \( \frac{d y}{d x}=e^{x+y-1} \)

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