Solve the following differential equation using using characteristic equation using laplace transform i. ii y" +y sin 2t, y(0) 2, y'(0) 1

The solution of the differential equation is

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

i) Using characteristic equation:

The characteristic equation method assumes that y(t)=, where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

As could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

Using Euler's formula:

The particular solution of the differential equation is given by:

So we use these derivatives in the differential equation:

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

Using the initial conditions we can check that C1=5/3 and C2=2

ii) Using Laplace Transform:

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]  

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]=

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) =

(s²+1)·Y(s)-2s-1=

(s²+1)·Y(s)=

Y(s)=

Y(s)=

Using partial franction method:

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)=

Y(s)=

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[]-ℒ⁻¹[]+ℒ⁻¹[]