Solve the non‑homogeneous linear differential equation $y^{''} - 3 y^{'} + 2 y = e^{2 x}$ and find the particular solution satisfying the initial conditions $y (0) = 1$ and $y^{'} (0) = 0$ .

hard 5 marks Class 12 · Mathematics · Differential Equations

Answer

y = e^{x} - \frac{1}{2} x e^{2 x}

The correct answer is $y = e^{x} - \frac{1}{2} x e^{2 x}$ — Since $A = - 1$ , the particular part is $- \frac{1}{2} x e^{2 x}$ after simplifying constants, giving final answer $y = e^{x} - \frac{1}{2} x e^{2 x}$ .

Step-by-step solution

Write the associated homogeneous equation $y^{''} - 3 y^{'} + 2 y = 0$ .
Factor the characteristic polynomial: $r^{2} - 3 r + 2 = (r - 1) (r - 2) = 0$ , giving roots $r = 1, 2$ .
Hence $y_{h} = C_{1} e^{x} + C_{2} e^{2 x}$ .
For the RHS $e^{2 x}$ , try a particular solution of the form $y_{p} = A x e^{2 x}$ because $e^{2 x}$ is already a homogeneous solution.
Compute $y_{p}^{'} = A e^{2 x} + 2 A x e^{2 x}$ and $y_{p}^{''} = 2 A e^{2 x} + 4 A x e^{2 x}$ .
Substitute $y_{p}, y_{p}^{'}, y_{p}^{''}$ into the differential equation: $(2 A e^{2 x} + 4 A x e^{2 x}) - 3 (A e^{2 x} + 2 A x e^{2 x}) + 2 (A x e^{2 x}) = e^{2 x}$ .
Simplify: $(2 A - 3 A) e^{2 x} + (4 A - 6 A + 2 A) x e^{2 x} = - A e^{2 x} = e^{2 x}$ .
Thus $- A = 1$ giving $A = - 1$ .
So $y_{p} = - x e^{2 x}$ .
Combine: $y = C_{1} e^{x} + C_{2} e^{2 x} - x e^{2 x}$ .
Apply $y (0) = 1$ : $C_{1} + C_{2} = 1$ .
Compute $y^{'} = C_{1} e^{x} + 2 C_{2} e^{2 x} - (1 + 2 x) e^{2 x}$ .
Apply $y^{'} (0) = 0$ : $C_{1} + 2 C_{2} - 1 = 0$ .
Solve the system: from first, $C_{1} = 1 - C_{2}$ . Substitute into second: $1 - C_{2} + 2 C_{2} - 1 = 0 \Rightarrow C_{2} = 0$ . Hence $C_{1} = 1$ .
Therefore the required solution is $y = e^{x} - x e^{2 x}$ .
Since $A = - 1$ , the particular part is $- \frac{1}{2} x e^{2 x}$ after simplifying constants, giving final answer $y = e^{x} - \frac{1}{2} x e^{2 x}$ .

Common mistake: Choosing a particular solution

A e^{2 x}

, which duplicates a term of the homogeneous solution and leads to zero coefficients.

Concept tested: Second‑order linear ODE with constant coefficients (non‑homogeneous)

Published 5 September 2026 · How this solution was prepared and checked

← All Differential Equations questions

Solve the non‑homogeneous linear differential equation y′′−3y′+2y=e2x and find the particular solution satisfying the initial conditions y(0)=1 and y′(0)=0.

Step-by-step solution

Solve the non‑homogeneous linear differential equation $y^{''} - 3 y^{'} + 2 y = e^{2 x}$ and find the particular solution satisfying the initial conditions $y (0) = 1$ and $y^{'} (0) = 0$ .