Consider the initial value problem below. The value of y at x = ln 2 , (rounded off to 3 decimal places ) is ______. $\frac{dy}{dx}=2x-y,y\left(0\right)=1$

To solve the given initial value problem, we need to find the specific solution $y(x)$ to the first-order linear differential equation and then evaluate it at $x=\ln 2$.

The correct answer is 0.886.

Step 1: Identify the type of differential equation
The given differential equation is:

$\frac{dy}{dx}=2x-y$

We can rewrite this in the standard form of a first-order linear differential equation, $\frac{dy}{dx}+P(x)y=Q(x)$:

$\frac{dy}{dx}+y=2x$

Here, $P(x)=1$ and $Q(x)=2x$.

Step 2: Find the Integrating Factor (I.F.)
The integrating factor $I.F.$ is calculated as:

$I.F.=e^{\int P(x)dx}=e^{\int 1dx}=e^x$

Step 3: Determine the general solution
Multiplying both sides of the standard equation by the integrating factor gives:

$y\cdot (I.F.)=\int Q(x)\cdot (I.F.)dx+C$

Substitute $I.F.=e^x$ and $Q(x)=2x$ into the equation:

$y{e}^x=\int 2x{e}^{x}dx+C$

Using integration by parts to evaluate $\int 2x{e}^{x}dx$:

$\int 2x{e}^{x}dx=2x{e}^x-\int 2e^{x}dx=2x{e}^x-2e^x$

Thus, the general solution is:

$y{e}^x=2x{e}^x-2e^x+C$

Dividing by $e^x$:

$y(x)=2x-2+C{e}^{-x}$

Step 4: Apply the initial condition
Using the given initial condition $y(0)=1$:

$1=2(0)-2+C{e}^0$

$1=-2+C\Rightarrow C=3$

So, the specific solution to the initial value problem is:

$y(x)=2x-2+3e^{-x}$

Step 5: Calculate $y$ at $x=\ln 2$
Substitute $x=\ln 2$ into the equation:

$y(\ln 2)=2\ln 2-2+3e^{-\ln 2}$

Since $e^{-\ln 2}=e^{\ln (1/2)}=\frac{1}{2}$:

$y(\ln 2)=2\ln 2-2+3\cdot \frac{1}{2}$

$y(\ln 2)=2\ln 2-2+1.5$

$y(\ln 2)=2\ln 2-0.5$

Using the value $\ln 2\approx 0.693147$:

$y(\ln 2)\approx 2(0.693147)-0.5$

$y(\ln 2)\approx 1.386294-0.5=0.886294$

Rounding off to 3 decimal places, we get 0.886.