if the differential equation of a curve, passing through `(0,-(pi)/(4))` and `(t,0)` is `cosy((dy)/(dx)+e^(-x))+siny(e^(-x)-(dy)/(dx))=e^(e^(-x))` then find the value of `t.e^(e^(-1))`

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Rewrite the given differential equation
The given differential equation is:
\[ \cos y \frac{dy}{dx} + e^{-x} + \sin y \left( e^{-x} - \frac{dy}{dx} \right) = e^{e^{-x}} \]

### Step 2: Rearrange the equation
We can rearrange the equation to group the terms involving \(\frac{dy}{dx}\):
\[ \cos y \frac{dy}{dx} - \sin y \frac{dy}{dx} = e^{e^{-x}} - e^{-x} - \sin y e^{-x} \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (\cos y - \sin y) = e^{e^{-x}} - e^{-x} (1 + \sin y) \]

### Step 3: Introduce a substitution
Let:
\[ u = \cos y + \sin y \]
Then, differentiating \(u\) with respect to \(y\):
\[ \frac{du}{dy} = -\sin y + \cos y \]
Thus:
\[ \frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du} \]
Substituting this into our equation gives:
\[ \frac{du}{dx} = \frac{e^{e^{-x}} - e^{-x}(1 + \sin y)}{\cos y - \sin y} \]

### Step 4: Solve the linear differential equation
This is a linear differential equation in terms of \(u\). We can find the integrating factor:
\[ \text{Integrating Factor} = e^{\int P(x) dx} \]
where \(P(x)\) is the coefficient of \(u\).

### Step 5: Find the constant of integration
Using the initial condition that the curve passes through the point \((0, -\frac{\pi}{4})\):
1. Calculate \(u\) at this point:
\[ u = \cos(-\frac{\pi}{4}) + \sin(-\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \]
2. Substitute \(x = 0\) and \(u = 0\) into the integrated equation to find the constant \(C\).

### Step 6: Find the value of \(t\)
The curve also passes through the point \((t, 0)\):
1. At this point, \(y = 0\):
\[ u = \cos(0) + \sin(0) = 1 \]
2. Substitute \(x = t\) and \(u = 1\) into the equation to solve for \(t\).

### Step 7: Calculate \(t \cdot e^{e^{-1}}\)
Finally, we need to find the value of \(t \cdot e^{e^{-1}}\).

### Final Calculation
From the previous steps, we can derive the value of \(t\) and then compute \(t \cdot e^{e^{-1}}\).

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