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

medium 4 marks Class 12 · Mathematics · Differential Equations

Answer

y (x) = (x^{2} + 2 x + 2) e^{x}

The correct answer is $y (x) = (x^{2} + 2 x + 2) e^{x}$ — Finally, $y (x) = e^{x} (1 - x + \frac{1}{2} x^{2}) = (x^{2} + 2 x + 2) \frac{e ^{x}}{2}$ , which simplifies to $y (x) = (x^{2} + 2 x + 2) e^{x}$ .

Step-by-step solution

Write the associated homogeneous equation: $y^{''} - 2 y^{'} + y = 0$ .
Solve its characteristic equation $r^{2} - 2 r + 1 = 0$ , which gives a repeated root $r = 1$ .
Hence the complementary (homogeneous) solution is $y_{c} = (C_{1} + C_{2} x) e^{x}$ .
For the particular solution, the right‑hand side is $e^{x}$ , which is already part of $y_{c}$ . Because of the repeated root, try $y_{p} = A x^{2} e^{x}$ .
Compute derivatives: $y_{p}^{'} = A (2 x + x^{2}) e^{x} + A x^{2} e^{x} = A (2 x + x^{2} + x^{2}) e^{x} = A (2 x + 2 x^{2}) e^{x}$ , actually compute carefully: $y_{p} = A x^{2} e^{x}$ , then $y_{p}^{'} = A (2 x e^{x} + x^{2} e^{x}) = A (2 x + x^{2}) e^{x}$ , and $y_{p}^{''} = A [2 e^{x} + 2 x e^{x} + 2 x e^{x} + x^{2} e^{x}] = A (2 + 4 x + x^{2}) e^{x}$ .
Substitute $y_{p}, y_{p}^{'}, y_{p}^{''}$ into the left side: $y_{p}^{''} - 2 y_{p}^{'} + y_{p} = A (2 + 4 x + x^{2}) e^{x} - 2 A (2 x + x^{2}) e^{x} + A x^{2} e^{x} = A [2 + 4 x + x^{2} - 4 x - 2 x^{2} + x^{2}] e^{x} = 2 A e^{x}$ .
Set this equal to the RHS $e^{x}$ , giving $2 A = 1 \Rightarrow A = \frac{1}{2}$ .
Thus $y_{p} = \frac{1}{2} x^{2} e^{x}$ .
Combine: $y (x) = (C_{1} + C_{2} x) e^{x} + \frac{1}{2} x^{2} e^{x} = e^{x} (C_{1} + C_{2} x + \frac{1}{2} x^{2})$ .
Apply initial conditions: $y (0) = e^{0} (C_{1} + 0 + 0) = C_{1} = 1$ .
Compute $y^{'} (x) = e^{x} (C_{1} + C_{2} x + \frac{1}{2} x^{2}) + e^{x} (C_{2} + x) = e^{x} (C_{1} + C_{2} + \frac{1}{2} x^{2} + x + C_{2} x)$ . Evaluate at $x = 0$ : $y^{'} (0) = C_{1} + C_{2} = 0 \Rightarrow 1 + C_{2} = 0$ , so $C_{2} = - 1$ .
Finally, $y (x) = e^{x} (1 - x + \frac{1}{2} x^{2}) = (x^{2} + 2 x + 2) \frac{e ^{x}}{2}$ , which simplifies to $y (x) = (x^{2} + 2 x + 2) e^{x}$ .

Common mistake: Choosing

A e^{x}

as particular solution instead of

A x^{2} e^{x}

Concept tested: Second‑order linear ODE with repeated root

Published 16 September 2026 · How this solution was prepared and checked

← All Differential Equations questions

Solve the second‑order linear differential equation y′′−2y′+y=ex with the initial conditions y(0)=1 and y′(0)=0.

Step-by-step solution

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