Question

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

Answer 1

The solution of the differential equation is
`y(t)= - (1)/(3) Sin(2t)+2 Cos(t)+(5)/(3) Sin(t)`

Explanation:

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

i) Using characteristic equation:

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

We find the solution of the homogeneus differential equation:

y" + y = 0

`y'=re^(rt)`

`y`

`r^(2)e^(rt)+e^(rt)=0`

`(r^(2)+1)e^(rt)=0`

As
`e^(rt)` could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

`y(t)_(h)=c_(1)e^(it)+c_(2)e^(-it)`

Using Euler's formula:

`y(t)_(h)=c_(1)[Sin(t)+iCos(t)]+c_(2)[Sin(t)-iCos(t)]`

`y(t)_(h)=(c_(1)+c_(2))Sin(t)+(c_(1)-c_(2))iCos(t)`

`y(t)_(h)=C_(1)Sin(t)+C_(2)Cos(t)`

The particular solution of the differential equation is given by:

`y(t)_(p)=ASin(2t)+BCos(2t)`

`y'(t)_(p)=2ACos(2t)-2BSin(2t)`

`y''(t)_(p)=-4ASin(2t)-4BCos(2t)`

So we use these derivatives in the differential equation:

`-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)`

`-3ASin(2t)-3BCos(2t)=Sin(2t)`

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:

`y(t)= - (1)/(3) Sin(2t) + C_(1) Sin(t) + C_(2) Cos(t)`

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)]=
`(2)/((s^(2)+4))`

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) =
`(2)/((s^(2)+4))`

(s²+1)·Y(s)-2s-1=
`(2)/((s^(2)+4))`

(s²+1)·Y(s)=
`(2)/((s^(2)+4))+2s+1=(2+2s(s^(2)+4)+s^(2)+4)/((s^(2)+4))`

Y(s)=
`(2+2s(s^(2)+4)+s^(2)+4)/((s^(2)+4)(s^(2)+1))`

Y(s)=
`(2s^(3)+s^(2)+8s+6)/((s^(2)+4)(s^(2)+1))`

Using partial franction method:

`(2s^(3)+s^(2)+8s+6)/((s^(2)+4)(s^(2)+1))=(As+B)/(s^(2)+4) +(Cs+D)/(s^(2)+1)`

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)=
`(-(2)/(3) )/(s^(2)+4) +(2s+(5)/(3) )/(s^(2)+1)`

Y(s)=
`-(1)/(3) (2)/(s^(2)+4) +2(s )/(s^(2)+1)+(5)/(3)(1)/(s^(2)+1)`

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[
`-(1)/(3) (2)/(s^(2)+4)`]-ℒ⁻¹[
`2(s )/(s^(2)+1)`]+ℒ⁻¹[
`(5)/(3)(1)/(s^(2)+1)`]

`y(t)= - (1)/(3) Sin(2t)+2 Cos(t)+(5)/(3) Sin(t)`