---
title: "Initial value problem y''−4y'+3y"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:initial-value-problem-y-4y"
source: https://www.bctengineer.com/q/semester_2/mathematics/initial-value-problem-y-4y
site: BCT Engineering
---

# Initial value problem y''−4y'+3y

Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions

- Asked in: 2075 Baisakh, 5 marks; 2068 Bhadra, 5 marks (asked 2 times)
- Video: [Solution to Second Order Differential Equation IVP | Initial Value Problem | 4y ′′ −4y ′ +3y=0](https://www.youtube.com/watch?v=dn6PhTsiDBc) — TEAM Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/initial-value-problem-y-4y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Solve the initial value problem: y'' − 4y' + 3y = 10e⁻²ˣ, y(0) = 1, y'(0) = 3.

## Short answer

Auxiliary roots are 1 and 3, so y_c = C_1 e^x + C_2 e^{3x}. Trial y_p = A e^{-2x} gives A = 2/3. Applying y(0)=1 and y'(0)=3 yields C_1 = -5/3, C_2 = 2. Final answer: y = -(5/3)e^x + 2e^{3x} + (2/3)e^{-2x}.

## Answer

Equation: y″ − 4y′ + 3y = 10e^(−2x), y(0) = 1, y′(0) = 3. A linear equation with constant coefficients: y = y_c + y_p.

#### Complementary function
Auxiliary equation m² − 4m + 3 = 0 gives (m − 1)(m − 3) = 0, so m = 1, 3.

y_c = C₁eˣ + C₂e^(3x).

#### Particular integral
Try y_p = Ae^(−2x). Then y_p′ = −2Ae^(−2x), y_p″ = 4Ae^(−2x):

```text
4A + 8A + 3A = 10   →   15A = 10   →   A = 2/3
```
(Equivalently y_p = 10e^(−2x)/f(−2), with f(D) = D² − 4D + 3, f(−2) = 4 + 8 + 3 = 15.)

General solution:

y = C₁eˣ + C₂e^(3x) + (2/3)e^(−2x)

#### Apply the initial conditions
y′ = C₁eˣ + 3C₂e^(3x) − (4/3)e^(−2x).

```text
y(0) = 1:   C₁ + C₂ + 2/3 = 1         →  C₁ + C₂ = 1/3
y′(0) = 3:  C₁ + 3C₂ − 4/3 = 3        →  C₁ + 3C₂ = 13/3
```
Subtract: 2C₂ = 4, so C₂ = 2, and C₁ = 1/3 − 2 = −5/3.

**Answer: y = −(5/3)eˣ + 2e^(3x) + (2/3)e^(−2x).**

Check: y(0) = −5/3 + 2 + 2/3 = 1 ✓; y′(0) = −5/3 + 6 − 4/3 = 3 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. m^2 - 4m + 3 = 0 gives:
   A. y_c = C1 e^(-x) + C2 e^(-3x)
   B. y_c = C1 e^x + C2 e^(3x)
   C. y_c = C1 e^x + C2 x e^x
   D. y_c = C1 e^(2x) + C2 e^(-2x)
   Answer: B. Roots m = 1, 3. (Key point: Complementary function)

2. For RHS 10 e^(-2x), y_p is:
   A. (2/3) e^(-2x)
   B. (10/3) e^(-2x)
   C. (2/3) e^(2x)
   D. (2/15) e^(-2x)
   Answer: A. 15A = 10. (Key point: Particular integral)

3. When are constants C1, C2 found?
   A. Before finding y_p
   B. Using only y(0)
   C. After forming the general solution, using y(0) and y'(0)
   D. By setting y_c = 0
   Answer: C. Apply the conditions to the full y = y_c + y_p. (Key point: Applying conditions)

4. The solution is:
   A. -(5/3)e^x + 2e^(3x) + (2/3)e^(-2x)
   B. (5/3)e^x - 2e^(3x) + (2/3)e^(-2x)
   C. -(5/3)e^x + 2e^(3x) - (2/3)e^(-2x)
   D. e^x + e^(3x)
   Answer: A. C2 = 2, C1 = -5/3. (Key point: Solution)

## Close variants set in other exams

- [Initial value problem y''+2y'+5y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Ainitial-value-problem-y-2y) — Mathematics, chapter 4, asked 3×

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