If (dy/dx ) = (2+lnx) ln has solution y = y(x) such that y(1) = e then y(e) is equal to

The correct answer is 3e – 1.

Step-by-step explanation:

We are given the first-order differential equation:
$\frac{dy}{dx}=2+\ln x$
with the initial condition:
$y\left(1\right)=e$

Step 1: Integrate the differential equation
We can find the general solution by integrating both sides of the equation with respect to $x$:
$y=\int (2+\ln x)dx$
We can split the integration into two parts:
$y=\int 2dx+\int \ln xdx$

Step 2: Evaluate the integrals
The first integral is simple:
$\int 2dx=2x$
For the second integral, we use integration by parts, $\int udv=uv-\int vdu$, where we let:
$u=\ln x\Rightarrow du=\frac{1}{x}dx$
and
$dv=dx\Rightarrow v=x$
Applying the integration by parts formula:
$\int \ln xdx=x\ln x-\int x\cdot \frac{1}{x}dx=x\ln x-x$

Combining the results of both integrals gives the general solution:
$y\left(x\right)=2x+x\ln x-x+C$
Simplifying this expression yields:
$y\left(x\right)=x\ln x+x+C$
where $C$ is the constant of integration.

Step 3: Solve for the constant C
Using the initial condition $y\left(1\right)=e$, we substitute $x=1$ and set the equation equal to $e$:
$e=1\cdot \ln \left(1\right)+1+C$
Since $\ln \left(1\right)=0$, we have:
$e=1+C$
Solving for $C$ gives:
$C=e-1$

Substituting $C$ back into our general solution gives the particular solution:
$y\left(x\right)=x\ln x+x+e-1$

Step 4: Find y(e)
Now we substitute $x=e$ into the particular solution:
$y\left(e\right)=e\ln \left(e\right)+e+e-1$
Since $\ln \left(e\right)=1$:
$y\left(e\right)=e\left(1\right)+e+e-1$
$y\left(e\right)=3e-1$