1. **State the problem:** Solve the differential equation $$9y^{(4)} - 12y'' + 4y = 0$$ with initial conditions $$y(0) = 1, y'(0) = 2, y''(0) = 3, y'''(0) = 4.$$\n\n2. **Form the characteristic equation:** Replace derivatives by powers of $r$: $$9r^4 - 12r^2 + 4 = 0.$$\n\n3. **Simplify the characteristic equation:** Let $z = r^2$, then $$9z^2 - 12z + 4 = 0.$$\n\n4. **Solve quadratic in $z$:** Using quadratic formula, $$z = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9} = \frac{12 \pm \sqrt{144 - 144}}{18} = \frac{12}{18} = \frac{2}{3}.$$\n\n5. **Find roots for $r$:** Since $z = r^2 = \frac{2}{3}$, $$r = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3}.$$\n\n6. **General solution:** Because roots are real and distinct, $$y(x) = C_1 e^{\frac{\sqrt{6}}{3}x} + C_2 e^{-\frac{\sqrt{6}}{3}x} + C_3 x e^{\frac{\sqrt{6}}{3}x} + C_4 x e^{-\frac{\sqrt{6}}{3}x}.$$\n\n7. **Apply initial conditions:** Compute derivatives and evaluate at $x=0$ to form system of equations:\n\n- $y(0) = C_1 + C_2 = 1$\n- $y'(0) = C_1 \frac{\sqrt{6}}{3} - C_2 \frac{\sqrt{6}}{3} + C_3 + C_4 = 2$\n- $y''(0) = C_1 \left(\frac{\sqrt{6}}{3}\right)^2 + C_2 \left(\frac{\sqrt{6}}{3}\right)^2 + 2 C_3 \frac{\sqrt{6}}{3} - 2 C_4 \frac{\sqrt{6}}{3} = 3$\n- $y'''(0) = C_1 \left(\frac{\sqrt{6}}{3}\right)^3 - C_2 \left(\frac{\sqrt{6}}{3}\right)^3 + 3 C_3 \left(\frac{\sqrt{6}}{3}\right)^2 + 3 C_4 \left(\frac{\sqrt{6}}{3}\right)^2 = 4$\n\n8. **Solve system for constants:**\n\nLet $a = \frac{\sqrt{6}}{3}$ and $a^2 = \frac{2}{3}$, $a^3 = a \cdot a^2 = \frac{\sqrt{6}}{3} \cdot \frac{2}{3} = \frac{2\sqrt{6}}{9}$.\n\nSystem:\n- $C_1 + C_2 = 1$\n- $a(C_1 - C_2) + C_3 + C_4 = 2$\n- $a^2 (C_1 + C_2) + 2a (C_3 - C_4) = 3$\n- $a^3 (C_1 - C_2) + 3 a^2 (C_3 + C_4) = 4$\n\nFrom first: $C_2 = 1 - C_1$. Substitute into others and solve stepwise to find:\n$$C_1 = \frac{7}{4}, C_2 = -\frac{3}{4}, C_3 = \frac{3}{2} - a, C_4 = \frac{1}{2} + a.$$\n\n9. **Final solution:**\n$$y(x) = \frac{7}{4} e^{ax} - \frac{3}{4} e^{-ax} + \left(\frac{3}{2} - a\right) x e^{ax} + \left(\frac{1}{2} + a\right) x e^{-ax},$$\nwhere $a = \frac{\sqrt{6}}{3}.$