Solve Difference Equation using Z-Transform | yₙ₊₂+6yₙ₊₁+9yₙ=2ⁿ | y₀=0, y₁=0 | BMATEC301 / BMATE301

🎓 Solve Difference Equation using Z-Transform | yₙ₊₂ + 6yₙ₊₁ + 9yₙ = 2ⁿ | BMATEC301 / BMATE301 | Mathematics-III In this video we solve the linear constant-coefficient difference equation using Z-Transform methods and given initial conditions: Given: yₙ₊₂ + 6yₙ₊₁ + 9yₙ = 2ⁿ , y₀=0, y₁=0 We apply the Z-transform to convert the recurrence to the z-domain, incorporate the initial conditions, solve for Y(z), and obtain y_n by taking the inverse Z-transform. The video shows full step-by-step algebra, partial fraction (or method of residues) where needed, and the final closed-form sequence — presented in VTU exam style suitable for BMATEC301 and BMATE301 students. 📚 Syllabus Reference – BMATEC301 / BMATE301 (Module 3 & 4): Z-Transforms and standard pairs Properties: Linearity, Shifting, Multiplication by n Application of initial conditions in z-domain Inverse Z-transform and sequence recovery Solving non-homogeneous difference equations with exponential inputs 🔍 Topics Covered: ✔ Forming Y(z) using Z-transform and given initial conditions ✔ Handling the forcing term 2^n in z-domain (use of standard Z-pair) ✔ Algebraic solution for Y(z) and partial fraction for inverse transform ✔ Final expression for y_n with VTU-style explanation and exam tips 💎 Support Us: Join our channel and get access to exclusive perks 👇 🔗    / @officialmathematicstutor   #DifferenceEquation #ZTransform #BMATEC301 #BMATE301 #MathematicsIII #VTUMathematics3 #VTUExamPreparation #VTUSyllabus #VTUModelPapers #VTUMaths3 #ECEngineering #EEEEngineering #InverseZTransform #ZTransformProblems