Solve the second‑order linear differential equation $y^{''} - 4 y^{'} + 5 y = e^{2 x}$ with the initial conditions $y (0) = 1$ and $y^{'} (0) = 0$ .

hard 4 marks Class 12 · Mathematics · Differential Equations

Answer

y = e^{2 x} (1 - 2 sin x)

The correct answer is $y = e^{2 x} (1 - 2 sin x)$ — $10. Substitute back: y = e^{2 x} (1 - 2 sin x)$

Step-by-step solution

$1. Solve the homogeneous part: r^{2} - 4 r + 5 = 0$
$2. Roots: r = 2 \pm i$ . Hence $y_{h} = e^{2 x} (C_{1} cos x + C_{2} sin x)$
$3. Guess a particular solution of the form y_{p} = A e^{2 x}$
$4. Substitute into the differential equation: 4 A - 8 A + 5 A = A$
$5. Set A = 1$ , so $y_{p} = e^{2 x}$
$6. General solution: y = e^{2 x} (C_{1} cos x + C_{2} sin x) + e^{2 x}$
$7. Apply y (0) = 1 : e^{0} (C_{1} + 0) + 1 = 1 \Rightarrow C_{1} = 0$
$8. Compute y^{'} :$ $y^{'} = e^{2 x} [2 C_{1} cos x + 2 C_{2} sin x + 2 + (- C_{1} sin x + C_{2} cos x)]$
$9. At x = 0 : y^{'} (0) = 2 C_{1} + 2 + C_{2} = 0 \Rightarrow 2 (0) + 2 + C_{2} = 0 \Rightarrow C_{2} = - 2$
$10. Substitute back: y = e^{2 x} (1 - 2 sin x)$

Common mistake: Forgetting to include the particular solution or making algebraic sign errors in constants

Concept tested: Second‑order linear differential equation