$2x_1+x_2+x_3+2x_4 = 54$. I know how to solve it when the coefficient of each variable is one but how to solve this kind of question?
What if I use this approach:
Let $y1 = 2x_1, y2 = x_2, y3 = x_3, y4 = 2x_4,$
$y_1+y_2+y_3+y_4 = 54$
Without restriction:
C(54+4-1,4-1) = C(57,3)
Subtract y1 or y4 is odd:
C(53+4-1, 4-1) = C(56,3)
Add back y1 and y4 are odd:
C(52+4-1, 4-1) = C(55,3)
So the answer is like: C(57,3)-2*C(56,3)+C(55,3) = 55
Is this approach valid?
You can try using generating functions, it should be the coefficient of $x^{54}$ in the function $$f(x) = \left(\sum_{n\geq 0} x^{2n} \right)^2\left(\sum_{n \geq 0}x^n\right)^2 = \frac{1}{(1-x)^2(1-x^2)^2}.$$
You should be able to get the exact answer by doing a partial fraction decomposition.
Your attempt is incorrect since the variables $y_1$ and $y_4$ can be odd. For instance, the ordered quadruple $(7, 9, 11, 10)$ satisfies the equation $2x_1 + x_2 + x_3 + 2x_4 = 54$, as do the ordered quadruples $(12, 5, 7, 9)$, and $(5, 13, 9, 11)$.
You evidently know that the number of solutions of the equation $$x_1 + x_2 + \cdots + x_k = n$$ in the nonnegative integers is given by the formula $$\binom{n + k - 1}{k - 1}$$
Since $x_2 + x_3 = 54 - 2x_1 - 2x_4$ is even, $x_2$ and $x_3$ must have the same parity, as Cameron Buie pointed out in the comments.
Let's consider cases.
Suppose $x_2$ and $x_3$ are both even. Then there exist nonnegative integers $y_2$ and $y_3$ such that $x_2 = 2y_2$ and $x_3 = 2y_3$. Substituting $2y_2$ for $x_2$ and $2y_3$ for $x_3$ in the equation $$2x_1 + x_2 + x_3 + 2x_4 = 54 \tag{1}$$ yields \begin{align*} 2x_1 + 2y_2 + 2y_3 + 2x_4 & = 54\\ x_1 + y_2 + y_3 + x_4 & = 27 \tag{2} \end{align*} Notice that equation $2$ is an equation in the nonnegative integers. Therefore, it has $$\binom{27 + 4 - 1}{4 - 1}$$ solutions.
Suppose $x_2$ and $x_3$ are both odd. Then there exist nonnegative integers $z_2$ and $z_3$ such that $x_2 = 2z_2 + 1$ and $x_3 = 2z_3 + 1$. Substituting $2z_2 + 1$ for $x_2$ and $2z_3 + 1$ for $x_3$ in equation $1$ yields \begin{align*} 2x_1 + 2z_2 + 1 + 2z_3 + 1 + 2x_4 & = 54\\ 2x_1 + 2z_2 + 2z_3 + x_4 & = 52\\ x_1 + z_2 + z_3 + x_4 & = 26 \tag{3} \end{align*} Notice that equation $3$ is also an equation in the nonnegative integers. Therefore, it has $$\binom{26 + 4 - 1}{4 - 1}$$ solutions.
Since the two cases are mutually exclusive and exhaustive, you can obtain the number of solutions of equation $1$ in the nonnegative integers by adding the number of solutions of equation $3$ to the number of solutions of equation $2$.
The generating function for the number of solutions in non-negative integers of $$2x_1+x_2+x_3+2x_4 = r$$ is $$f(x) = \frac{1}{(1-x)^2 (1-x^2)^2}$$ We want $[x^{54}]f(x)$, the coefficient of $x^{54}$ when $f(x)$ is expanded as an infinite series. Note that $$\begin{align} f(x) &= \frac{(1-x^2)^2}{(1-x)^2} \cdot \frac{1}{(1-x^2)^4} \\ &= (1+x)^2 \cdot \frac{1}{(1-x^2)^4} \\ &= (1+2x+x^2) \cdot (1-x^2)^{-4} \\ &= (1+2x+x^2) \cdot \sum_{i=0}^{\infty} \binom{4+i-1}{i} x^{2i} \end{align}$$ by the Binomial Therem for negative exponents. So $$\begin{align} [x^{54}]f(x) &= [x^{54}](1+2x+x^2) \cdot \sum_{i=0}^{\infty} \binom{4+i-1}{i} x^{2i} \\ &= [x^{54}] \sum_{i=0}^{\infty} \binom{4+i-1}{i} x^{2i} + 2[x^{53}]\sum_{i=0}^{\infty} \binom{4+i-1}{i} x^{2i} \\ & \qquad + [x^{52}]\sum_{i=0}^{\infty} \binom{4+i-1}{i} x^{2i} \\ &= \binom{4+27-1}{27} + 2 \cdot 0 +\binom{4+26-1}{26} \\ &= 7714 \end{align}$$
