1. **Problem Statement:** We are given the Dequan-Li system of ordinary differential equations (ODEs): $$\dot{x} = a(y - x) + cxz$$ $$\dot{y} = 55x + 20y - xz$$ $$\dot{z} = bz + xy - 0.65 x^2$$ with parameters $a=40$, $b=1.833$, and $c=0.16$. We want to find the fixed points and analyze their stability. 2. **Finding Fixed Points:** Fixed points occur where the derivatives are zero: $$\dot{x} = 0, \quad \dot{y} = 0, \quad \dot{z} = 0.$$ So we solve the system: $$0 = a(y - x) + cxz,$$ $$0 = 55x + 20y - xz,$$ $$0 = bz + xy - 0.65 x^2.$$ 3. **Rewrite the system:** From the first equation: $$a(y - x) + cxz = 0 \implies a y - a x + c x z = 0 \implies a y = a x - c x z = x(a - c z).$$ So, $$y = \frac{x(a - c z)}{a}.$$ 4. **Substitute $y$ into the second equation:** $$0 = 55 x + 20 y - x z = 55 x + 20 \cdot \frac{x(a - c z)}{a} - x z = x \left(55 + \frac{20(a - c z)}{a} - z \right).$$ Since $x$ could be zero or not, consider two cases: - Case 1: $x=0$ - Case 2: $x \neq 0$, then $$55 + \frac{20(a - c z)}{a} - z = 0.$$ Multiply both sides by $a$: $$55 a + 20(a - c z) - a z = 0 \implies 55 a + 20 a - 20 c z - a z = 0 \implies 75 a = z (20 c + a).$$ So, $$z = \frac{75 a}{20 c + a}.$$ 5. **Substitute $y$ and $z$ into the third equation:** Recall: $$0 = b z + x y - 0.65 x^2.$$ Substitute $y = \frac{x(a - c z)}{a}$: $$0 = b z + x \cdot \frac{x(a - c z)}{a} - 0.65 x^2 = b z + \frac{x^2 (a - c z)}{a} - 0.65 x^2.$$ Rewrite: $$b z + x^2 \left( \frac{a - c z}{a} - 0.65 \right) = 0.$$ 6. **Analyze Case 1: $x=0$** Then from step 3, $y = \frac{0}{a} = 0$. From step 5: $$0 = b z + 0 - 0 = b z \implies z = 0.$$ So one fixed point is: $$(x,y,z) = (0,0,0).$$ 7. **Analyze Case 2: $x \neq 0$** We have $z = \frac{75 a}{20 c + a}$ from step 4. Calculate $a - c z$: $$a - c z = a - c \cdot \frac{75 a}{20 c + a} = a \left(1 - \frac{75 c}{20 c + a} \right) = a \cdot \frac{20 c + a - 75 c}{20 c + a} = a \cdot \frac{a - 55 c}{20 c + a}.$$ Substitute into step 5: $$0 = b z + x^2 \left( \frac{a - c z}{a} - 0.65 \right) = b z + x^2 \left( \frac{a - c z - 0.65 a}{a} \right).$$ Simplify numerator: $$a - c z - 0.65 a = (a - 0.65 a) - c z = 0.35 a - c z.$$ Recall $c z = c \cdot \frac{75 a}{20 c + a}$, so $$0.35 a - c z = 0.35 a - \frac{75 a c}{20 c + a} = a \left(0.35 - \frac{75 c}{20 c + a} \right).$$ Therefore, $$0 = b z + x^2 \cdot \frac{a \left(0.35 - \frac{75 c}{20 c + a} \right)}{a} = b z + x^2 \left(0.35 - \frac{75 c}{20 c + a} \right).$$ Rearranged: $$x^2 = - \frac{b z}{0.35 - \frac{75 c}{20 c + a}}.$$ 8. **Plug in parameter values:** $$a=40, b=1.833, c=0.16.$$ Calculate denominator: $$20 c + a = 20 \times 0.16 + 40 = 3.2 + 40 = 43.2,$$ $$\frac{75 c}{20 c + a} = \frac{75 \times 0.16}{43.2} = \frac{12}{43.2} \approx 0.2778,$$ $$0.35 - 0.2778 = 0.0722.$$ Calculate $z$: $$z = \frac{75 a}{20 c + a} = \frac{75 \times 40}{43.2} = \frac{3000}{43.2} \approx 69.44.$$ Calculate $x^2$: $$x^2 = - \frac{b z}{0.0722} = - \frac{1.833 \times 69.44}{0.0722} = - \frac{127.3}{0.0722} \approx -1763.5.$$ Since $x^2$ is negative, no real $x$ exists for this case. 9. **Conclusion:** The only real fixed point is at the origin: $$(x,y,z) = (0,0,0).$$ 10. **Stability Analysis:** The Jacobian matrix $J$ at a point $(x,y,z)$ is: $$J = \begin{bmatrix} - a + c z & a & c x \\ 55 - z & 20 & -x \\ y - 1.3 x & x & b \end{bmatrix}.$$ At $(0,0,0)$: $$J = \begin{bmatrix} -40 & 40 & 0 \\ 55 & 20 & 0 \\ 0 & 0 & 1.833 \end{bmatrix}.$$ The eigenvalues determine stability. The $z$-component eigenvalue is $1.833 > 0$, indicating instability. **Final answer:** The Dequan-Li system with given parameters has a single real fixed point at the origin $(0,0,0)$, which is unstable due to a positive eigenvalue in the Jacobian matrix.