Let's solve the given differential equation for the beam deflection: ### Given Differential Equation \[ EI\frac{d^4y}{dx^4} = w(x) \] where \( w(x) = w_0 \) (a constant load), \( EI \) is the flexural rigidity, and the beam length is \( L \). ### Step 1: Substitute \( w(x) \) \[ EI\frac{d^4y}{dx^4} = w_0 \] \[ \frac{d^4y}{dx^4} = \frac{w_0}{EI} \] ### Step 2: Integrate Four Times Integrate both sides four times with respect to \( x \): #### First integration: \[ \frac{d^3y}{dx^3} = \frac{w_0}{EI}x + C_1 \] #### Second integration: \[ \frac{d^2y}{dx^2} = \frac{w_0}{2EI}x^2 + C_1x + C_2 \] #### Third integration: \[ \frac{dy}{dx} = \frac{w_0}{6EI}x^3 + \frac{C_1}{2}x^2 + C_2x + C_3 \] #### Fourth integration: \[ y(x) = \frac{w_0}{24EI}x^4 + \frac{C_1}{6}x^3 + \frac{C_2}{2}x^2 + C_3x + C_4 \] ### Step 3: Boundary Conditions The beam is: - **Embedded at left end \(x=0\)**: Both displacement and slope are zero. - \( y(0) = 0 \) - \( y'(0) = 0 \) - **Simply supported at right end \(x=L\)**: Displacement and bending moment are zero. - \( y(L) = 0 \) - \( y''(L) = 0 \) ### Step 4: Apply Boundary Conditions 1. At \( x = 0 \): - \( y(0) = 0 \implies C_4 = 0 \) - \( y'(0) = 0 \implies C_3 = 0 \) 2. At \( x = L \): - \( y(L) = 0 \): \[ \frac{w_0}{24EI}L^4 + \frac{C_1}{6}L^3 + \frac{C_2}{2}L^2 = 0 \] - \( y''(L) = 0 \): \[ \frac{w_0}{2EI}L^2 + C_1L + C_2 = 0 \] Now solve the above two equations for \( C_1 \) and \( C_2 \): From \( y''(L) = 0 \): \[ \frac{w_0}{2EI}L^2 + C_1L + C_2 = 0 \implies C_2 = -\frac{w_0}{2EI}L^2 - C_1L \] Plug into \( y(L) = 0 \): \[ \frac{w_0}{24EI}L^4 + \frac{C_1}{6}L^3 + \frac{1}{2}\left(-\frac{w_0}{2EI}L^2 - C_1L\right)L^2 = 0 \] \[ \frac{w_0}{24EI}L^4 + \frac{C_1}{6}L^3 - \frac{w_0}{4EI}L^4 - \frac{C_1}{2}L^3 = 0 \] Combine like terms: \[ \left(\frac{w_0}{24EI}L^4 - \frac{w_0}{4EI}L^4\right) + \left(\frac{C_1}{6}L^3 - \frac{C_1}{2}L^3\right) = 0 \] \[ -\frac{w_0}{6EI}L^4 - \frac{C_1}{3}L^3 = 0 \] \[ -\frac{w_0}{6EI}L^4 = \frac{C_1}{3}L^3 \implies C_1 = -\frac{w_0}{2EI}L \] Now \( C_2 \): \[ C_2 = -\frac{w_0}{2EI}L^2 - C_1L = -\frac{w_0}{2EI}L^2 - \left(-\frac{w_0}{2EI}L\right)L = -\frac{w_0}{2EI}L^2 + \frac{w_0}{2EI}L^2 = 0 \] ### Step 5: Write Final Solution \[ y(x) = \frac{w_0}{24EI}x^4 - \frac{w_0}{12EI}L x^3 \] **Final Answer:** \[ \boxed{ y(x) = \frac{w_0}{24EI}\left(x^4 - 2Lx^3\right) } \] Or, equivalently, \[ \boxed{ y(x) = \frac{w_0}{24EI} \left[ x^4 - 2Lx^3 \right] } \] --- Let me know if you need further explanation or have follow-up questions!

The solution involves integrating the differential equation four times to obtain the deflection equation, applying boundary conditions to determine integration constants, and ensuring the beam's displacement and moments are satisfied at the supports. This method provides the beam's deflection profile under uniform load with specified support and boundary conditions.