Question

solve the problem d^2x/dt^2 +x=8cos2t-4sint, x(π/2)=-1, x'(π/2)=0

Answer 1

Solution

Generally ,the Laplace transform of a derivative is given by;

L{fⁿ}=sⁿF(s)-s^n-1f(0)-s^n-2f'(0)....f^n-1(0)

Since the given boundary conditions don't give the x(0) and x'(0),

Take x(0)=A and x'(0)=B

The Laplace transform of the equation will be written as;

s²F(s)-sA-B+ F(s)=8cos2t-4sint

We know Laplace transform of cos at and sin at from the tables

Therefore;

(s²+1)F(s)-sA-B=$\frac{8s}{s^2+2^2}-\frac{4}{s^2+1^2}$

Make F(s) the subject of the formula;

F(s)=$\frac{8s}{(s^2+4)(s^2+1)}$ -$\frac{4}{(s^2+1)(s^2+1)}$+$\frac{sA}{s^2+1}$ +$\frac{B}{s^2+1}$

Using convolution method,it gives

x(t)=8cos2tsint-4sin²t+Acost+Bsint

x'(t)=8costcos2t-16sintsin2t-8costsint-Asint+Bcost

Apply the given conditions to find A and B

x(π/2)=-1

-1=8(-1)(1)-4(1)+A(0)+B(1)

B=11

x'(π/2)=0

0=8(0)(-1)-16(1)(0)-8(0)(1)-A(1)+B(0)

A=0

Therefore the particular solution is;

x(t)=8cos2tsint-4sin²t+11sint