Solve
$$\begin{cases}
\dfrac{\mathrm dx(t)}{\mathrm dt}&=-y(t),\\\\
\dfrac{\mathrm dy(t)}{\mathrm dt}&=\dfrac{101}{100}x(t)-y(t);
\end{cases}$$
for $\{x(t),\,y(t)\}$.
Integrate $\frac{\mathrm dx(t)}{\mathrm dt}=-y(t)$ with respect to $t$:
$$\int\frac{\mathrm dx(t)}{\mathrm dt}\,\mathrm dt=\int\Big(-y(t)\Big)\,\mathrm dt;$$
evaluate the integrals:
$$x(t)=-\int y(t)\,\mathrm dt;$$
substitute $x(t)=-\int y(t)\,\mathrm dt$ in $\frac{\mathrm dy(t)}{\mathrm dt}=\frac{101}{100}x(t)-y(t)$:
$$\frac{\mathrm dy(t)}{\mathrm dt}=-\frac{101}{100}\int y(t)\,\mathrm dt-y(t);$$
substitute $u(t)=\int y(t)\,\mathrm dt \implies y(t)=\frac{\mathrm du(t)}{\mathrm dt}$:
$$\frac{\mathrm d}{\mathrm dt}\frac{\mathrm du(t)}{\mathrm dt}=-\frac{101}{100}\int \frac{\mathrm du(t)}{\mathrm dt}\,\mathrm dt-\frac{\mathrm du(t)}{\mathrm dt};$$
simplify:
$$\frac{\mathrm d^2u(t)}{\mathrm dt^2}=-\frac{101}{100}u(t)-\frac{\mathrm du(t)}{\mathrm dt};$$
rearrange the equation:
$$\frac{\mathrm d^2u(t)}{\mathrm dt^2}+\frac{\mathrm du(t)}{\mathrm dt}+\frac{101}{100}u(t)=0\tag{1}\label{ode1}$$
for the homogeneous linear second-order ordinary differential equation $\eqref{ode1}$, substitute $u(t)=e^{\lambda t}$:
$$\frac{\mathrm d^2e^{\lambda t}}{\mathrm dt^2}+\frac{\mathrm de^{\lambda t}}{\mathrm dt}+\frac{101}{100}e^{\lambda t}=0;$$
evaluate the differentials:
$$\lambda^2e^{\lambda t}+\lambda e^{\lambda t}+\frac{101}{100}e^{\lambda t}=0;$$
factor out $e^{\lambda t}$:
$$e^{\lambda t}\left(\lambda^2+\lambda+\frac{101}{100}\right)=0;$$
since for any finite $\lambda$ and $t$: $e^{\lambda t}\neq0$, the roots must come from the polynomial $\lambda^2+\lambda+\frac{101}{100}$:
$$\lambda^2+\lambda+\frac{101}{100}=0;$$
multiply both sides by $4$:
$$(2\lambda)^2+4\lambda+\frac{404}{100}=0;$$
rewrite $\frac{404}{100}=1+\frac{304}{100}$:
$$(2\lambda)^2+4\lambda+1+\frac{304}{100}=0;$$
complete the square:
$$(2\lambda+1)^2+\frac{304}{100}=0;$$
isolate the term $(2\lambda+1)^2$:
$$(2\lambda+1)^2=-\frac{304}{100};$$
take the square roots of both sides:
$$2\lambda+1=\pm\frac{2i}{5}\sqrt{19};$$
isolate the term $2\lambda$:
$$2\lambda=-1\pm\frac{2i}{5}\sqrt{19};$$
divide both sides by two:
$$\lambda=\frac{-1\pm\frac{2i}{5}\sqrt{19}}{2};$$
then the solutions to $\eqref{ode1}$ are given by:
$$u(t)=c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}+c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t},$$
where $c_1$ and $c_2$ are arbitrary constants and $i$ is the imaginary unit, such that $i^2=-1$; substitute back $u(t)=\int y(t)\,\mathrm dt$:
$$\int y(t)\,\mathrm dt=c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}+c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t};$$
differentiate both sides with respect to $t$:
$$\frac{\mathrm d}{\mathrm dt}\int y(t)\,\mathrm dt=\frac{\mathrm d}{\mathrm dt}\left(c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}+c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}\right);$$
evaluate the differentials:
$$y(t)=\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t};$$
substitute for $y(t)$ in $\frac{\mathrm dx(t)}{\mathrm dt}=-y(t)$:
$$\frac{\mathrm dx(t)}{\mathrm dt}=-\left(\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}\right);$$
integrate both sides with respect to $t$:
$$\int\frac{\mathrm dx(t)}{\mathrm dt}\mathrm dt=-\int\left(\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}\right)\mathrm dt;$$
evaluate the integrals:
$$x(t)=-c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}+c_3,$$
where $c_3$ is an arbitrary constant; collect the solutions:
$$\therefore\begin{cases}
x(t)=\boxed{c_3-c_1e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-c_2e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}}\phantom{.},\\\\
y(t)=\boxed{\frac{c_1}{2}\left(2i\sqrt{19}/5-1\right)e^{\frac{1}{2}\left(2i\sqrt{19}/5-1\right)t}-\frac{c_2}{2}\left(1+2i\sqrt{19}/5\right)e^{\frac{1}{2}\left(-1-2i\sqrt{19}/5\right)t}}\phantom{.}.
\end{cases}$$
I think that the process you're envisioning, if you did it rigorously, becomes equivalent to Picard's iterative process: Given an initial value problem $y'(t)= f(t,y)$, $y(0) = y_0$, the sequence of functions defined by
\begin{align*}
\phi_0(t) &= y_0 \\
\phi_{k+1}(t) &= y_0 + \int_0^t f(s,\phi_k(s))\,ds
\end{align*}
converges to a solution $\phi(t)$.
(You can think of that integral as summing $f(s,\phi_k(s))$ at many discrete points between $s=0$ and $s=t$, multiplied by a small amount $\Delta s$, and taking the limit as $\Delta s \to 0$.)
In your problem, you have $f(t,y) = y$ and $y_0 = 1$. Picard's process gives:
\begin{align*}
\phi_0(t) &= 1 \\
\phi_1(t) &= 1 + \int_0^t \phi_0(s)\,ds = 1 + \int_0^t 1\,ds = 1 + t \\
\phi_2(t) &= 1 + \int_0^t(1+s)\,ds = 1 + t + \tfrac{1}{2}t^2 \\
\phi_3(t) &= 1 + \int_0^t\left(1 + s + \tfrac{1}{2}s^2\right)\,ds = 1 + t + \tfrac{1}{2}t^2 + \tfrac{1}{6} t^3
\end{align*}
You can see that $\phi_k(t)$ is the $k$-th degree Taylor polynomial for $\phi(t) = e^t$, and that is the solution to the IVP.
Best Answer
Let's consider the initial condition
$$y(0)=y_0.$$
Then
$$y'(0)=F(0,y_0)=-0.04 \sqrt{y_0}.$$
Then assuming a time step $\Delta t=h$ we can estimate
$$y(h)\approx y_1= y(0)+y'(0)\cdot \Delta t=y_0+y'_0\cdot h \implies F(h,y_1)=-0.04 \sqrt{y_1}.$$
Then
$$y(2h)\approx y_2=y_1+h\cdot F(h,y_0) \implies F(2h,y_2)=-0.04 \sqrt{y_2}$$
and so on.
Refer also to