Let's solve the given ODE in **two steps** with explanations after each step, and then summarize the final answer. --- ## 1. **Theoretical (Analytical) Solution** The equation is: \[ 2\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = \cos x - 5\sin x \] ### **Step 1: Solve the Homogeneous Equation** The homogeneous equation is: \[ 2y'' + 2y' + 5y = 0 \] Divide both sides by 2: \[ y'' + y' + \frac{5}{2}y = 0 \] **Characteristic equation:** \[ r^2 + r + \frac{5}{2} = 0 \] Solve for \( r \): \[ r = \frac{-1 \pm \sqrt{1 - 4 \times \frac{5}{2}}}{2} = \frac{-1 \pm \sqrt{1 - 10}}{2} = \frac{-1 \pm \sqrt{-9}}{2} = \frac{-1 \pm 3i}{2} \] So, \[ r_1 = -\frac{1}{2} + \frac{3}{2}i, \quad r_2 = -\frac{1}{2} - \frac{3}{2}i \] **Homogeneous solution:** \[ y_h(x) = e^{-\frac{1}{2}x}(C_1 \cos(\frac{3}{2}x) + C_2 \sin(\frac{3}{2}x)) \] **Explanation:** We solved the homogeneous part using the characteristic equation, resulting in complex roots. This gives us a decaying oscillatory solution. --- ### **Step 2: Find a Particular Solution** Guess a particular solution of the form: \[ y_p = A\cos x + B\sin x \] Compute derivatives: \[ y_p' = -A\sin x + B\cos x \] \[ y_p'' = -A\cos x - B\sin x \] Plug into the original equation: \[ 2(-A\cos x - B\sin x) + 2(-A\sin x + B\cos x) + 5(A\cos x + B\sin x) = \cos x - 5\sin x \] Expand: \[ (-2A + 2B + 5A)\cos x + (-2B - 2A + 5B)\sin x = \cos x - 5\sin x \] \[ (3A + 2B)\cos x + (3B - 2A)\sin x = \cos x - 5\sin x \] Set coefficients equal: - For \(\cos x\): \(3A + 2B = 1\) - For \(\sin x\): \(3B - 2A = -5\) Solve these equations: 1. \(3A + 2B = 1\) 2. \(-2A + 3B = -5\) Multiply first equation by 2: \(6A + 4B = 2\) Add to second equation multiplied by 3: \(-6A + 9B = -15\) Add: \((6A + 4B) + (-6A + 9B) = 2 + (-15)\) \(13B = -13 \implies B = -1\) Plug \(B = -1\) into first equation: \(3A + 2(-1) = 1 \implies 3A - 2 = 1 \implies 3A = 3 \implies A = 1\) So, \[ y_p(x) = \cos x - \sin x \] **Explanation:** We guessed a particular solution, substituted, and solved for \(A\) and \(B\). This gives a steady-state response matching the nonhomogeneous part. --- ### **General Solution** \[ \boxed{ y(x) = e^{-\frac{1}{2}x}(C_1 \cos(\frac{3}{2}x) + C_2 \sin(\frac{3}{2}x)) + \cos x - \sin x } \] --- ## 2. **MATLAB Solution** You can use `dsolve` in MATLAB to solve this ODE symbolically. ```matlab syms y(x) ode = 2*diff(y,x,2) + 2*diff(y,x) + 5*y == cos(x) - 5*sin(x); sol = dsolve(ode) ``` This will output the same general solution as above. **Explanation:** MATLAB's `dsolve` function computes the general solution symbolically, confirming the analytical approach. --- ### **Final Answer** \[ \boxed{ y(x) = e^{-\frac{1}{2}x}(C_1 \cos(\frac{3}{2}x) + C_2 \sin(\frac{3}{2}x)) + \cos x - \sin x } \]