Given differential equation: \[ 2x^2 y'' + 4x y' + 3y = 0 \] Let's solve it step by step. --- ### **Step 1: Recognize the Type** This is a **Cauchy-Euler (Euler-Cauchy) equation** of the form: \[ a x^2 y'' + b x y' + c y = 0 \] --- ### **Step 2: Substitute \( y = x^r \)** Let \( y = x^r \). Then, \[ y' = r x^{r-1} \] \[ y'' = r(r-1)x^{r-2} \] Substitute these into the equation: \[ 2x^2 [r(r-1)x^{r-2}] + 4x [r x^{r-1}] + 3[x^r] = 0 \] \[ 2r(r-1)x^{r} + 4r x^{r} + 3 x^r = 0 \] \[ [2r(r-1) + 4r + 3] x^r = 0 \] --- ### **Step 3: Characteristic Equation** Factor out \(x^r\) (which is not zero): \[ 2r(r-1) + 4r + 3 = 0 \] \[ 2r^2 - 2r + 4r + 3 = 0 \] \[ 2r^2 + 2r + 3 = 0 \] --- ### **Step 4: Solve the Quadratic** \[ 2r^2 + 2r + 3 = 0 \] Use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 2\), \(b = 2\), \(c = 3\): \[ r = \frac{-2 \pm \sqrt{2^2 - 4\cdot2\cdot3}}{2\cdot2} \] \[ r = \frac{-2 \pm \sqrt{4 - 24}}{4} \] \[ r = \frac{-2 \pm \sqrt{-20}}{4} \] \[ r = \frac{-2 \pm 2i\sqrt{5}}{4} \] \[ r = \frac{-1 \pm i\sqrt{5}}{2} \] --- ### **Step 5: General Solution** For complex roots \( r = \alpha \pm i\beta \), the general solution is: \[ y(x) = x^{\alpha}\left[ C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x) \right] \] Here, \( \alpha = -\frac{1}{2} \), \( \beta = \frac{\sqrt{5}}{2} \): \[ \boxed{ y(x) = x^{-1/2} \left[ C_1 \cos\left( \frac{\sqrt{5}}{2} \ln x \right) + C_2 \sin\left( \frac{\sqrt{5}}{2} \ln x \right) \right] } \] --- **Final Answer:** \[ \boxed{ y(x) = x^{-1/2} \left[ C_1 \cos\left( \frac{\sqrt{5}}{2} \ln x \right) + C_2 \sin\left( \frac{\sqrt{5}}{2} \ln x \right) \right] } \]