Math Problem Statement

solve

Solve the differential equation y'' + y = sin(2t), y(0) = 0, y'(0) = 1 using Laplace transforms.

Solution

Let's solve the given problem step by step from the Laplace transform part onward.

Problem:

The differential equation is: y+y=sin(2t),y(0)=0,y(0)=1y'' + y = \sin(2t), \quad y(0) = 0, \quad y'(0) = 1

We already have the Laplace transform taken in the uploaded file:

s2Y(s)sy(0)y(0)+Y(s)=2s2+4s^2Y(s) - sy(0) - y'(0) + Y(s) = \frac{2}{s^2 + 4}

Substituting the initial conditions y(0)=0y(0) = 0 and y(0)=1y'(0) = 1, the equation simplifies to: s2Y(s)1+Y(s)=2s2+4s^2Y(s) - 1 + Y(s) = \frac{2}{s^2 + 4}

Reorganizing: (s2+1)Y(s)=1+2s2+4(s^2 + 1)Y(s) = 1 + \frac{2}{s^2 + 4}

Thus: Y(s)=1s2+1+2(s2+4)(s2+1)Y(s) = \frac{1}{s^2 + 1} + \frac{2}{(s^2 + 4)(s^2 + 1)}

Step 1: Take the inverse Laplace transform term by term

First term:

1s2+1\frac{1}{s^2 + 1} This is the Laplace transform of sin(t)\sin(t): L1(1s2+1)=sin(t)\mathcal{L}^{-1}\left(\frac{1}{s^2 + 1}\right) = \sin(t)

Second term:

2(s2+4)(s2+1)\frac{2}{(s^2 + 4)(s^2 + 1)} We need to use partial fraction decomposition to split this into simpler terms.

Let: 2(s2+4)(s2+1)=As2+1+Bs+Cs2+4\frac{2}{(s^2 + 4)(s^2 + 1)} = \frac{A}{s^2 + 1} + \frac{Bs + C}{s^2 + 4}

Multiply through by (s2+4)(s2+1)(s^2 + 4)(s^2 + 1): 2=A(s2+4)+(Bs+C)(s2+1)2 = A(s^2 + 4) + (Bs + C)(s^2 + 1)

Expanding both terms: 2=A(s2)+4A+Bs(s2)+Bs+Cs2+C2 = A(s^2) + 4A + Bs(s^2) + Bs + Cs^2 + C 2=As2+4A+Bs3+Bs+Cs2+C2 = A s^2 + 4A + B s^3 + Bs + Cs^2 + C

Group terms by powers of ss: 2=Bs3+(A+C)s2+Bs+(4A+C)2 = B s^3 + (A + C)s^2 + Bs + (4A + C)

Equating coefficients with 2=0s3+0s2+0s+22 = 0s^3 + 0s^2 + 0s + 2:

  • Coefficient of s3s^3: B=0B = 0
  • Coefficient of s2s^2: A+C=0C=AA + C = 0 \Rightarrow C = -A
  • Coefficient of ss: B=0B = 0 (already known)
  • Constant term: 4A+C=24A + C = 2

Substitute C=AC = -A into 4A+C=24A + C = 2: 4AA=23A=2A=23,C=234A - A = 2 \Rightarrow 3A = 2 \Rightarrow A = \frac{2}{3}, \, C = -\frac{2}{3}

Thus, the partial fraction decomposition is: 2(s2+4)(s2+1)=23s2+123s2+4\frac{2}{(s^2 + 4)(s^2 + 1)} = \frac{\frac{2}{3}}{s^2 + 1} - \frac{\frac{2}{3}}{s^2 + 4}

Now, take the inverse Laplace transform of each term.

  1. 23s2+1\frac{\frac{2}{3}}{s^2 + 1} is 23sin(t)\frac{2}{3} \sin(t),
  2. 23s2+4-\frac{\frac{2}{3}}{s^2 + 4} is 23sin(2t)/2=13sin(2t)-\frac{2}{3} \sin(2t)/2 = -\frac{1}{3} \sin(2t).

Step 2: Combine the results

The solution y(t)y(t) is: y(t)=sin(t)+23sin(t)13sin(2t)y(t) = \sin(t) + \frac{2}{3}\sin(t) - \frac{1}{3}\sin(2t)

Simplify: y(t)=(1+23)sin(t)13sin(2t)y(t) = \left(1 + \frac{2}{3}\right)\sin(t) - \frac{1}{3}\sin(2t) y(t)=53sin(t)13sin(2t)y(t) = \frac{5}{3}\sin(t) - \frac{1}{3}\sin(2t)

Final Answer:

y(t)=53sin(t)13sin(2t)y(t) = \frac{5}{3}\sin(t) - \frac{1}{3}\sin(2t)


Would you like me to explain any specific step in greater detail? Let me know!

Here are 5 related questions you might find helpful:

  1. What is the Laplace transform and how is it used to solve differential equations?
  2. How do we perform partial fraction decomposition for more complex fractions?
  3. How does the inverse Laplace transform work for trigonometric functions?
  4. Can we solve this differential equation using a different method, like undetermined coefficients?
  5. How does changing the initial conditions affect the solution?

Tip: Always double-check your partial fraction decomposition by substituting it back into the original equation to ensure correctness!

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Laplace Transform
Inverse Laplace Transform
Partial Fraction Decomposition
Trigonometric Functions

Formulas

Laplace Transform: \(\mathcal{L}\{f(t)\} = F(s)\)
Inverse Laplace Transform: \(\mathcal{L}^{-1}\{F(s)\} = f(t)\)
Partial Fraction Decomposition

Theorems

Superposition Principle for Linear Systems
Laplace Transform Theorem for Derivatives

Suitable Grade Level

Undergraduate Mathematics or Advanced High School (Grade 12)