The solution of the differential equation `(dy)/(dx) + (y)/(x log_(e)x)=(1)/(x)` under the condition y=1 when x=e is

Options

A. `2y= log_(e)x + (1)/(log_(e)x)`
B. `y= log_(e)x + (2)/(log_(e)x)`
C. `y log_(e) x= log_(e)x +1`
D. `y= log_(e) x +e`

Answer

To solve the differential equation

\[ \frac{dy}{dx} + \frac{y}{x \log_e x} = \frac{1}{x} \]

with the condition \( y = 1 \) when \( x = e \), we will follow these steps:

### Step 1: Identify the form of the differential equation

The given equation is in the standard linear form:

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

where \( P(x) = \frac{1}{x \log_e x} \) and \( Q(x) = \frac{1}{x} \).

### Step 2: Find the integrating factor

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

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x \log_e x} \, dx} \]

To solve this integral, we can use the substitution \( t = \log_e x \), which implies \( dt = \frac{1}{x} dx \) or \( dx = x dt = e^t dt \). Therefore, we have:

\[ \int \frac{1}{x \log_e x} \, dx = \int \frac{1}{t} \, dt = \log_e |t| + C = \log_e |\log_e x| + C \]

Thus, the integrating factor becomes:

\[ \mu(x) = e^{\log_e |\log_e x|} = \log_e x \]

### Step 3: Multiply the differential equation by the integrating factor

Now, we multiply the entire differential equation by \( \log_e x \):

\[ \log_e x \frac{dy}{dx} + \frac{y}{x} = \frac{\log_e x}{x} \]

### Step 4: Rewrite the left-hand side as a derivative

The left-hand side can be rewritten as:

\[ \frac{d}{dx}(y \log_e x) = \frac{\log_e x}{x} \]

### Step 5: Integrate both sides

Integrating both sides gives:

\[ y \log_e x = \int \frac{\log_e x}{x} \, dx \]

The integral on the right can be solved using integration by parts. Let \( u = \log_e x \) and \( dv = \frac{1}{x} dx \). Then \( du = \frac{1}{x} dx \) and \( v = \log_e x \). Thus,

\[ \int \frac{\log_e x}{x} \, dx = \frac{(\log_e x)^2}{2} + C \]

So we have:

\[ y \log_e x = \frac{(\log_e x)^2}{2} + C \]

### Step 6: Solve for \( y \)

Now, we can solve for \( y \):

\[ y = \frac{(\log_e x)^2}{2 \log_e x} + \frac{C}{\log_e x} = \frac{\log_e x}{2} + \frac{C}{\log_e x} \]

### Step 7: Apply the initial condition

Now we apply the initial condition \( y = 1 \) when \( x = e \):

\[ 1 = \frac{\log_e e}{2} + \frac{C}{\log_e e} \]

Since \( \log_e e = 1 \):

\[ 1 = \frac{1}{2} + C \implies C = \frac{1}{2} \]

### Step 8: Substitute back the value of \( C \)

Substituting \( C \) back into the equation for \( y \):

\[ y = \frac{\log_e x}{2} + \frac{1/2}{\log_e x} \]

### Final Solution

Thus, the solution of the differential equation is:

\[ y = \frac{\log_e x}{2} + \frac{1}{2 \log_e x} \]