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

Based on the provided image, we can identify the differential equation as:
$\frac{dy}{dx}+4y=5$
which is valid in the domain $0\le x\le 1$ with the initial condition $y\left(0\right)=2.25$.

To solve this first-order linear ordinary differential equation, we write it in the standard form:
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
By comparing, we have:
$P\left(x\right)=4$
and
$Q\left(x\right)=5$

The integrating factor (I.F.) is given by:
$I.F.=e^{\int P\left(x\right)dx}=e^{\int 4dx}=e^{4x}$

Multiplying both sides of the differential equation by the integrating factor:
$e^{4x}\frac{dy}{dx}+4e^{4x}y=5e^{4x}$
This simplifies to:
$\frac{d}{dx}\left(y{e}^{4x}\right)=5e^{4x}$

Integrating both sides with respect to $x$:
$y{e}^{4x}=\int 5e^{4x}dx$
$y{e}^{4x}=\frac{5}{4}$
where $C$ is the constant of integration.

Dividing both sides by $e^{4x}$ yields:
$y=1.25+C{e}^{-4x}$

Next, we apply the initial condition $y\left(0\right)=2.25$ to find $C$:
$2.25=1.25+C{e}^{-4\left(0\right)}$
$2.25=1.25+C\left(1\right)$
$C=2.25-1.25=1$

Substituting $C=1$ back into the general solution gives the particular solution:
$y=e^{-4x}+1.25$