The differential equation dy is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of the differential equation is

The correct option is y = e^-4x + 1.25.

Step 1: Identify the differential equation from the image
Based on the provided image, the differential equation is:
$$\frac{dy}{dx}+4y=5$$
This is a first-order linear ordinary differential equation of the standard form:
$$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$$
Comparing the equations, we identify:
$$P\left(x\right)=4$$
$$Q\left(x\right)=5$$

Step 2: Determine the Integrating Factor (I.F.)
To solve this type of differential equation, we find the integrating factor:
$$\text{I.F.}=e^{\int P\left(x\right)dx}=e^{\int 4dx}=e^{4x}$$

Step 3: Solve the General differential Equation
The general solution for a first-order linear differential equation is given by:
$$y\cdot \left(\text{I.F.}\right)=\int Q\left(x\right)\cdot \left(\text{I.F.}\right)dx+C$$
Substituting our terms into the solution formula:
$$y\cdot e^{4x}=\int 5e^{4x}dx+C$$
Integrating the right-hand side gives:
$$y\cdot e^{4x}=\frac{5}{4}{e}^{4x}+C$$
where $C$ is the constant of integration. Dividing both sides by $e^{4x}$ yields the general solution:
$$y=1.25+C{e}^{-4x}$$

Step 4: Apply the boundary condition to find C
We are given that $y\left(0\right)=2.25$. Plugging in $x=0$ and $y=2.25$:
$$2.25=1.25+C{e}^0$$
Since $e^0=1$:
$$2.25=1.25+C$$
Subtracting $1.25$ from both sides:
$$C=1$$

Step 5: Write the final particular solution
Substituting $C=1$ back into the general solution gives:
$$y=e^{-4x}+1.25$$
Thus, the final solution is verified as correct.