A question regarding Runge-Kutta method of order 3. 
Runge-Kutta method is defined by the iterations:


$u_{i+1}=u_i+ahf(x_i,u_i)+bhf(x_i+ch,u_i+dhf(x_
i,u_i))$


where $a,b,c,d$ are constants.
Given that the method converges of order 3.  Show that $a+b=1$, $cb=1/2$ , $db=1/2$

It is quite not useful i guess to use the definition of this method which appears in wikipedia page:
https://www.google.com/url.
How can one show this in another way, not based on the definition in the link above.I thought about using the Taylor expansion twice and compare coefficients but I am not sure how to do it correctly.
 A: [Note] : The idea that you should be aware of is that a $k^{th}$ order Runge-Kutta method scheme approximates $y(t_{i+1})$ by $y_{i+1}$ recursively with a global error of $O(h^{k})$ by using $k$ function evaluations. I will present you the derivation for Runge-Kutta method of order $2$ so that you get to derive to use similar argument to derive the third order Runge-Kutta method.
Therefore, we shall derive the Runge-Kutta method of order $2$ that is of the form :
$$
\left\{\begin{array}{l}
K_{1}=f\left(t_{i}, y_{i}\right) \\
K_{2}=f\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right) \\
y_{i+1}=y_{i}+h\left(w_{1} K_{1}+w_{2} K_{2}\right)
\end{array}\right.
$$
Where $\alpha,\beta,w_{1}$, and $w_{2}$ are constants that we will determine.
The scheme is obtained by assuming that the exact value $y\left(t_{i}+h\right)$ satisfies the equation $(1)$. Then, $y\left(t_{i}\right)$ is approximated by $y_{i},$ which leads to equation $(2)$
\begin{align*}
y\left(t_{i}+h\right) &=y\left(t_{i}\right)+w_{1} h f\left(t_{i},y\left(t_{i}\right)\right)+w_{2} h f\left(t_{i}+\alpha h,y\left(t_{i}\right)+\beta h f\left(t_{i}, y\left(t_{i}\right)\right)\right)+O\left(h^{3}\right) \tag1\\
&=y_{i}+w_{1} h K_{1}+w_{2} h f\left(t_{i}+\alpha h, y_{i}+\beta hK_{1}\right)+O\left(h^{3}\right) \tag2
\end{align*}
where $\alpha, \beta, w_{1},$ and $w_{2}$ relations are obtained by comparing $(2)$ with the Taylor series expansion of $y(t)$ at the point $t_{i}$ for
$t=t_{i}+h$ up to second order.
$$
\begin{align*}
y\left(t_{i}+h\right) &=y\left(t_{i}\right)+h y^{\prime}\left(t_{i}\right)+0.5 h^{2} y^{\prime \prime}\left(t_{i}\right)+O\left(h^{3}\right) \\
y^{\prime}\left(t_{i}\right) &=f\left(t_{i}, y\left(t_{i}\right)\right) \\
y^{\prime \prime}(t) &=\left(y^{\prime}(t)\right)^{\prime}=\frac{d}{d t} f(t, y(t))=f_{t}(t, y(t))+y^{\prime}(t) f_{y}(t, y(t)) \quad \text {(chain rule)} \\
&=f_{t}(t, y(t))+f(t, y(t)) f_{y}(t, y(t)) \\
y^{\prime \prime}\left(t_{i}\right) &=f_{t}\left(t_{i}, y\left(t_{i}\right)\right)+f\left(t_{i}, y\left(t_{i}\right)\right) f_{y}\left(t_{i}, y\left(t_{i}\right)\right) \\
\Longrightarrow\left(t_{i}+h\right) &=y\left(t_{i}\right)+h f\left(t_{i}, y\left(t_{i}\right)\right)+\frac{h^{2}}{2}\left[f_{t}\left(t_{i}, y\left(t_{i}\right)\right)+f\left(t_{i}, y\left(t_{i}\right)\right) f_{y}\left(t_{i}, y\left(t_{i}\right)\right)\right]+O\left(h^{3}\right) \\
&=y_{i}+h K_{1}+\frac{h^{2}}{2}\left[f_{t}\left(t_{i}, y_{i}\right)+K_{1} f_{y}\left(t_{i}, y_{i}\right)\right]+O\left(h^{3}\right)\tag3
\end{align*}
$$
[Note] : Till now we can't just compare $(3)$ and $(2)$ because of the term $f(t_{i}+\alpha h,y_{i}+\beta h K_{1})$. So using Taylor series expansion
for the function $\phi(h):=f(t_{i}+\alpha h,y_{i}+\beta h K_{1})$ centered at zero up to first order, and then replacing it in $(2)$ we get :
$$
\begin{align*}
\phi(h) &=\phi(0)+h \phi^{\prime}(0)+O\left(h^{2}\right)=f\left(t_{i}, y_{i}\right)+h\left(\frac{d}{d h} f\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right)\right)(0)+O\left(h^{2}\right) \\
&=f\left(t_{i}, y_{i}\right)+h\left(\alpha f_{t}\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right)+\beta K_{1} f_{y}\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right)\right)(0)+O\left(h^{2}\right) \\
&=K_{1}+\alpha h f_{t}\left(t_{i}, y_{i}\right)+\beta h K_{1} f_{y}\left(t_{i}, y_{i}\right)+O\left(h^{2}\right) \\
\Longrightarrow y\left(t_{i}+h\right) &=y_{i}+w_{1} h K_{1}+w_{2} h\left[K_{1}+\alpha h f_{t}\left(t_{i}, y_{i}\right)+\beta h K_{1} f_{y}\left(t_{i}, y_{i}\right)+O\left(h^{2}\right)\right]+O\left(h^{3}\right) \\
&=y_{i}+\left(w_{1}+w_{2}\right) h K_{1}+h^{2} w_{2}\left[\alpha f_{t}\left(t_{i}, y_{i}\right)+\beta K_{1} f_{y}\left(t_{i}, y_{i}\right)\right]+O\left(h^{3}\right)\tag4
\end{align*}
$$
By comparing equation (4) to (3), we get:
$$
\left\{\begin{array}{ll}
w_{1}+w_{2}=1 & \\
w_{2} \alpha=\frac{1}{2} & \text { where } \\
w_{2} \beta=\frac{1}{2}
\end{array} \quad\left\{\begin{array}{l}
K_{1}=f\left(t_{i}, y_{i}\right) \\
K_{2}=f\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right) \\
y_{i+1}=y_{i}+h\left(w_{1} K_{1}+w_{2} K_{2}\right)
\end{array}\right.\right.
$$
As desired.
[Final Note] : Since we have $3$ equations and $4$ unknowns, then we can obtain infinitely many second order RK schemes. For instance let $w_{1}=w_{2}=\frac{1}{2}$, then $\alpha=\beta=1$ and then you would have deduced a special type of RK2 scheme :)
Now try deriving the RK3 scheme using this approach and you should get the result you want.
