Let `y'(x)+y(x)g'(x)=g(x) g'(x) , y(0)=0, x in R` where f'(x) denotes `(d(f(x)))/(dx)` and g(x) is a given non-constant differentiable function of R with g(0)=g(2)=0. Then y(2) is equal to

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

To solve the differential equation given by

\[ y'(x) + y(x) g'(x) = g(x) g'(x), \quad y(0) = 0, \]

we can approach it using an integrating factor method.

### Step 1: Identify the Integrating Factor

The equation is in the standard form of a first-order linear differential equation:

\[ y' + P(x)y = Q(x), \]

where \( P(x) = g'(x) \) and \( Q(x) = g(x)g'(x) \).

The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int g'(x) \, dx} = e^{g(x)}. \]

### Step 2: Multiply the Equation by the Integrating Factor

Now, we multiply the entire differential equation by the integrating factor:

\[ e^{g(x)} y' + e^{g(x)} y g'(x) = e^{g(x)} g(x) g'(x). \]

### Step 3: Rewrite the Left Side as a Derivative

The left side can be rewritten as the derivative of a product:

\[ \frac{d}{dx}(e^{g(x)} y) = e^{g(x)} g(x) g'(x). \]

### Step 4: Integrate Both Sides

Now we integrate both sides with respect to \( x \):

\[ \int \frac{d}{dx}(e^{g(x)} y) \, dx = \int e^{g(x)} g(x) g'(x) \, dx. \]

The left side simplifies to:

\[ e^{g(x)} y = \int e^{g(x)} g(x) g'(x) \, dx + C, \]

where \( C \) is the constant of integration.

### Step 5: Solve the Right Side Integral

To solve the right side, we can use substitution. Let \( u = g(x) \), then \( du = g'(x) dx \). The integral becomes:

\[ \int e^u u \, du. \]

Using integration by parts, let \( v = u \) and \( dw = e^u du \):

- \( dv = du \)
- \( w = e^u \)

Then,

\[ \int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u + C. \]

So,

\[ \int e^{g(x)} g(x) g'(x) \, dx = e^{g(x)} g(x) - e^{g(x)} + C. \]

### Step 6: Substitute Back

Substituting back, we have:

\[ e^{g(x)} y = e^{g(x)} g(x) - e^{g(x)} + C. \]

Dividing through by \( e^{g(x)} \):

\[ y = g(x) - 1 + Ce^{-g(x)}. \]

### Step 7: Apply Initial Condition

Using the initial condition \( y(0) = 0 \):

\[ 0 = g(0) - 1 + Ce^{-g(0)}. \]

Since \( g(0) = 0 \):

\[ 0 = 0 - 1 + C \cdot 1 \Rightarrow C = 1. \]

### Step 8: Final Solution

Thus, the solution becomes:

\[ y = g(x) - 1 + e^{-g(x)}. \]

### Step 9: Evaluate at \( x = 2 \)

Now we need to find \( y(2) \):

\[ y(2) = g(2) - 1 + e^{-g(2)}. \]

Since \( g(2) = 0 \):

\[ y(2) = 0 - 1 + e^{0} = -1 + 1 = 0. \]

### Final Answer

Thus, \( y(2) = 0 \).

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