Let's solve **part (a) theoretically** in 3 steps, with clear explanations after each step. --- ### **(a) Theoretical Solution** #### **Step 1: Write the Characteristic Equation** Given: \[ \frac{d^2y}{dt^2} - 6\frac{dy}{dt} + 10y = 0 \] Assume \(y = e^{rt}\): Substitute into the equation: \[ r^2e^{rt} - 6re^{rt} + 10e^{rt} = 0 \implies r^2 - 6r + 10 = 0 \] **Explanation:** We convert the differential equation into an algebraic equation by assuming a solution of the form \(e^{rt}\). This leads to the characteristic equation. --- #### **Step 2: Solve the Characteristic Equation** \[ r^2 - 6r + 10 = 0 \] Use the quadratic formula: \[ r = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 10}}{2} = \frac{6 \pm \sqrt{36 - 40}}{2} = \frac{6 \pm \sqrt{-4}}{2} = \frac{6 \pm 2i}{2} = 3 \pm i \] **Explanation:** The roots are complex (\(3+i\) and \(3-i\)), so the solution involves exponentials and sines/cosines. --- #### **Step 3: Write the General Solution and Solve for Constants** General solution: \[ y(t) = e^{3t}(A\cos t + B\sin t) \] Apply initial conditions: - \(y(0) = 1\): \[ y(0) = e^{0}(A\cos 0 + B\sin 0) = A = 1 \] - \(y'(t) = e^{3t}[(A\cos t + B\sin t)'] + 3e^{3t}(A\cos t + B\sin t)\) \[ y'(t) = e^{3t}(-A\sin t + B\cos t) + 3e^{3t}(A\cos t + B\sin t) \] At \(t=0\): \[ y'(0) = (0 + B) + 3A = B + 3A = 0 \] Since \(A=1\): \(y'(0) = B + 3 \times 1 = 0 \implies B = -3\) **Explanation:** We apply the initial conditions to solve for constants \(A\) and \(B\). --- #### **Final Answer (a):** \[ \boxed{ y(t) = e^{3t}(\cos t - 3\sin t) } \] --- Would you like the MATLAB code next, or should I solve part (b) theoretically first?

Let's proceed with solving both parts step-by-step. --- ## Part (a): Theoretical Solution ### Step 1: Write the characteristic equation Given differential equation: \[ \frac{d^2 y}{dt^2} - 6 \frac{dy}{dt} + 10 y = 0 \] Assuming solutions of the form \( y = e^{rt} \): \[ r^2 - 6r + 10 = 0 \] **Explanation:** Replacing \( y = e^{rt} \) transforms the differential equation into an algebraic quadratic, enabling us to find the roots which determine the solution form. --- ### Step 2: Find roots of the characteristic equation Calculate discriminant: \[ \Delta = (-6)^2 - 4 \times 1 \times 10 = 36 - 40 = -4 \] Since \(\Delta < 0\), roots are complex conjugates: \[ r = \frac{6 \pm \sqrt{-4}}{2} = \frac{6 \pm 2i}{2} = 3 \pm i \] **Explanation:** Complex roots indicate oscillatory solutions combined with exponential growth/decay. --- ### Step 3: Write the general solution and find particular constants General solution: \[ y(t) = e^{3t} \left( C_1 \cos t + C_2 \sin t \right) \] Given initial conditions: - \( y(0) = 1 \): \[ y(0) = e^{0} (C_1 \times 1 + C_2 \times 0) = C_1 = 1 \] - \( y'(t) \): \[ y'(t) = e^{3t} \left[ 3(C_1 \cos t + C_2 \sin t) + (-C_1 \sin t + C_2 \cos t) \right] \] At \( t=0 \): \[ y'(0) = e^{0} [ 3 C_1 + C_2 ] = 3 C_1 + C_2 \] Given \( y'(0) = 0 \): \[ 3 \times 1 + C_2 = 0 \Rightarrow C_2 = -3 \] **Final solution:** \[ \boxed{ \boxed{ y(t) = e^{3t} (\cos t - 3 \sin t) } } \] --- ## Part (b): Find the general solution for: \[ x^2 y'' + 2x y' + y = \cos x - S_5 \sin x \] ### Step 1: Recognize the type of differential equation This is a Cauchy-Euler (equidimensional) differential equation: \[ x^2 y'' + 2x y' + y = \text{RHS} \] The homogeneous part: \[ x^2 y'' + 2x y' + y = 0 \] has characteristic roots \( r \) from assuming \( y = x^r \). --- ### Step 2: Solve the homogeneous equation Substitute \( y = x^r \): \[ r(r-1) x^{r-2} + 2 r x^{r-2} + x^{r} = 0 \] Dividing through by \( x^{r-2} \): \[ r(r-1) + 2 r + 1 = 0 \] Simplify: \[ r^2 - r + 2 r + 1 = 0 \Rightarrow r^2 + r + 1 = 0 \] Solve quadratic: \[ r = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i \sqrt{3}}{2} \] Thus, the homogeneous solution: \[ y_h = x^{-1/2} \left[ C_1 \cos \left( \frac{\sqrt{3}}{2} \ln x \right) + C_2 \sin \left( \frac{\sqrt{3}}{2} \ln x \right) \right] \] ### Step 3: Find a particular solution Right side: \( \cos x - S_5 \sin x \) (assuming \( S_5 \) is a constant coefficient). Since RHS are standard trigonometric functions, look for particular solutions: \[ y_p = A \cos x + B \sin x \] Differentiate: \[ y_p' = -A \sin x + B \cos x \] \[ y_p'' = -A \cos x - B \sin x \] Substitute into original: \[ x^2 (-A \cos x - B \sin x) + 2x (-A \sin x + B \cos x) + (A \cos x + B \sin x) = \cos x - S_5 \sin x \] This is complicated because of the \( x^2 \) term; a better approach is to use variation of parameters or numerical methods, but since the RHS is standard, and the homogeneous solution involves powers of \( x \), the particular solution will be more complex. **In conclusion:** - The **general solution** is the sum of the homogeneous solution (as above) and a particular solution, which can be found via variation of parameters or advanced methods. --- ## **Final answers:** ### **(a) Solution:** \[ \boxed{ y(t) = e^{3t} (\cos t - 3 \sin t) } \] ### **(b) General solution:** \[ \boxed{ y(x) = x^{-\frac{1}{2}} \left[ C_1 \cos \left( \frac{\sqrt{3}}{2} \ln x \right) + C_2 \sin \left( \frac{\sqrt{3}}{2} \ln x \right) \right] + y_p } \] where \( y_p \) is a particular solution to the nonhomogeneous equation involving \( \cos x - S_5 \sin x \). --- If you'd like, I can also provide MATLAB code for solving these equations numerically or symbolically. Would you like to proceed with that?

Let's solve **part (b)** step-by-step, providing the theoretical solution process and explanation. --- ### **Part (b): Find the general solution of** \[ x^2 y'' + 2x y' + y = \cos x - S_5 \sin x \] --- ### **Step 1: Recognize the type of differential equation** This is a **Cauchy-Euler (equidimensional)** differential equation: \[ x^2 y'' + 2x y' + y = \text{RHS} \] **Explanation:** Cauchy-Euler equations have variable coefficients that are powers of \( x \). Their homogeneous solutions are typically found by assuming solutions of the form \( y = x^r \). --- ### **Step 2: Find the homogeneous solution** Assuming \( y_h = x^r \), compute derivatives: \[ y' = r x^{r-1} \] \[ y'' = r(r-1) x^{r-2} \] Substitute into the homogeneous part: \[ x^2 \times r(r-1) x^{r-2} + 2x \times r x^{r-1} + x^r = 0 \] Simplify: \[ r(r-1) x^{r} + 2r x^{r} + x^{r} = 0 \] Factor \( x^{r} \): \[ x^{r} [ r(r-1) + 2r + 1 ] = 0 \] Set the bracketed term to zero: \[ r(r-1) + 2r + 1 = 0 \] \[ r^2 - r + 2r + 1 = 0 \] \[ r^2 + r + 1 = 0 \] Solve quadratic: \[ r = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \] --- ### **Step 3: Write the homogeneous solution** Since roots are complex conjugates: \[ r = \frac{-1}{2} \pm i \frac{\sqrt{3}}{2} \] The homogeneous solution: \[ y_h = x^{\alpha} \left[ C_1 \cos \left( \beta \ln x \right) + C_2 \sin \left( \beta \ln x \right) \right] \] where: \[ \alpha = -\frac{1}{2}, \quad \beta = \frac{\sqrt{3}}{2} \] **Therefore:** \[ \boxed{ \textbf{Homogeneous solution:} \quad y_h = x^{-1/2} \left[ C_1 \cos \left( \frac{\sqrt{3}}{2} \ln x \right) + C_2 \sin \left( \frac{\sqrt{3}}{2} \ln x \right) \right] } \] --- ### **Step 4: Find a particular solution \( y_p \)** Right side: \[ f(x) = \cos x - S_5 \sin x \] Since RHS involves standard trigonometric functions and the homogeneous solution involves powers of \( x \) and logs, we can attempt a particular solution of the form: \[ y_p = A \cos x + B \sin x \] **But** because the differential operator involves powers of \( x \), and the RHS is standard trig functions, the method of variation of parameters is better suited here. **In practice:** - The particular solution can be found using **variation of parameters** with the homogeneous solutions. - The method involves computing the Wronskian of the homogeneous solutions and integrating accordingly. --- ### **Final form:** \[ \boxed{ \textbf{General solution:} \quad y(x) = y_h + y_p } \] where: - \( y_h \) is as above. - \( y_p \) is the particular solution involving integrals (via variation of parameters). --- ### **Summary:** - The **homogeneous solution** involves powers and logs: \[ y_h = x^{-1/2} \left[ C_1 \cos \left( \frac{\sqrt{3}}{2} \ln x \right) + C_2 \sin \left( \frac{\sqrt{3}}{2} \ln x \right) \right] \] - The **particular solution** corresponds to solving a nonhomogeneous equation with RHS \( \cos x - S_5 \sin x \), typically via variation of parameters or undetermined coefficients (more advanced methods). --- **Would you like to see the MATLAB implementation for this, or further details on the particular solution?**