---
title: "Solve ODE by Laplace transform y''+4y'+3y=t"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-14"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-14
site: BCT Engineering
---

# Solve ODE by Laplace transform y''+4y'+3y=t

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2068 Chaitra, 5 marks
- Video: [Linear differential equation using Laplace transform example(PART-3) by easy maths easy tricks](https://www.youtube.com/watch?v=IE1DKTsJL3A) — EASY MATHS EASY TRICKS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-14
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Solve the equation using Laplace transform y'' + 4y' + 3y = t, t > 0, y(0) = 0, y'(0) = 1.

## Short answer

Apply Laplace transform, partial fractions, invert: y(t) = t/3 - 4/9 + e^(-t) - (5/9)e^(-3t).

## Answer

y″ + 4y′ + 3y = t, y(0) = 0, y′(0) = 1.

**Step 1: Transform** (Y = L[y]).

```text
(s²Y − 0 − 1) + 4(sY − 0) + 3Y = 1/s²
(s² + 4s + 3) Y = 1 + 1/s² = (s² + 1)/s²
Y = (s² + 1) / [ s² (s + 1)(s + 3) ]
```

**Step 2: Partial fractions.**

```text
Y = A/s + B/s² + C/(s+1) + D/(s+3)
B = 1/(1·3) = 1/3
C (s = −1) = (1 + 1)/(1·2) = 1
D (s = −3) = (9 + 1)/(9·(−2)) = −5/9
A: A + C + D = 0 (large s) → A = −4/9
```

**Step 3: Invert.**

**Answer: y(t) = t/3 − 4/9 + e^(−t) − (5/9) e^(−3t)**

**Check:** y(0) = −4/9 + 1 − 5/9 = 0 ✓; y′(0) = 1/3 − 1 + 5/3 = 1 ✓.

## Check yourself

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

1. With y(0) = 0 and y'(0) = 1, what is L[y'']?
   A. s^2 Y - 1
   B. s^2 Y
   C. s^2 Y + 1
   D. s Y - 1
   Answer: A. L[y''] = s^2 Y - s y(0) - y'(0) = s^2 Y - 1. (Key point: Transform of y'')

2. After transforming y'' + 4y' + 3y = t, what is (s^2 + 4s + 3)Y equal to?
   A. 1/s^2
   B. 1 + 1/s^2
   C. 1 - 1/s^2
   D. 1 + 1/s
   Answer: B. The initial condition adds 1 and L[t] = 1/s^2 is the right side. (Key point: Algebraic equation)

3. For Y = (s^2+1)/[s^2(s+1)(s+3)], what is the coefficient C of 1/(s+1)?
   A. 1/2
   B. 2
   C. 1
   D. -1
   Answer: C. At s = -1: (1+1)/(1*2) = 1. (Key point: Partial fraction constant C)

4. What is y(t)?
   A. t/3 + 4/9 + e^(-t) - (5/9)e^(-3t)
   B. t/3 - 4/9 + e^(-t) + (5/9)e^(-3t)
   C. t/3 - 4/9 - e^(-t) - (5/9)e^(-3t)
   D. t/3 - 4/9 + e^(-t) - (5/9)e^(-3t)
   Answer: D. Inverting each term gives this; it satisfies y(0) = 0 and y'(0) = 1. (Key point: Final solution)

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## Related in other subjects

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