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Step-by-Step Solution
Step 1: Write Down the Differential Equation
The given differential equation is:
$ e^y \left( \frac{dy}{dx} - 1 \right) = e^x $
Step 2: Expand and Rearrange the Equation
Distribute $ e^y :$
$ e^y \frac{dy}{dx} - e^y = e^x. $
Rewriting gives:
$ e^y \frac{dy}{dx} = e^x + e^y. $
Step 3: Substitute $ t = e^y $
Let $ t = e^y. $ Then, note that
$ \frac{dt}{dx} = e^y \frac{dy}{dx}. $
So the differential equation becomes:
$ \frac{dt}{dx} - t = e^x. $
(We rewrote $e^y \frac{dy}{dx} - e^y = e^x$ in the form $ \frac{dt}{dx} - t = e^x.)
Step 4: Recognize and Solve the First-Order Linear Differential Equation in $t$
The standard form of a first-order linear differential equation is:
$ \frac{dt}{dx} + P(x)\,t = Q(x). $
Here, $P(x) = -1$ and $Q(x) = e^x,$ giving:
$ \frac{dt}{dx} - t = e^x.
The integrating factor (IF) is given by:
$ \text{IF} = e^{\int P(x)\,dx} = e^{\int -1 \,dx} = e^{-x}. $
Multiplying both sides of the differential equation by this integrating factor, we get:
$ e^{-x} \frac{dt}{dx} - e^{-x} t = e^x \cdot e^{-x}.
Notice that the left-hand side becomes the derivative of $t e^{-x},$ so we can write:
$ \frac{d}{dx} \bigl( t e^{-x} \bigr) = 1.
Step 5: Integrate Both Sides
Integrating with respect to $x$:
$ t e^{-x} = \int 1 \, dx = x + C,
where $C$ is the constant of integration.
Step 6: Convert Back to $ y $
Since $t = e^y,$ we have:
$ e^y \, e^{-x} = x + C
\quad \Longrightarrow \quad
e^{y - x} = x + C.
This implies:
$ y - x = \ln(x + C).
Hence the general solution can be written as:
$ y = x + \ln(x + C).
Step 7: Use the Initial Condition $ y(0) = 0 $
We are given $ y(0) = 0. $ Substitute $ x = 0 $ and $ y = 0 $ into the solution:
$ 0 = 0 + \ln(0 + C)
\quad \Longrightarrow \quad
\ln(C) = 0
\quad \Longrightarrow \quad
C = 1.
Step 8: Write the Particular Solution
With $ C = 1, $ our solution becomes:
$ y = x + \ln(x + 1).
Step 9: Find $ y(1) $
Substitute $ x = 1 $ into the particular solution:
$ y(1) = 1 + \ln(1 + 1) = 1 + \ln(2).
This matches the given correct answer.
Final Answer:
$ y(1) = 1 + \ln(2).
