How to solve for third order differential equation of y"'-7y'+6y =2 sin (t) using Method of Laplace Transform when y(0)=0, y'(0)=0, y"(0)=0? Step by step

bobbie71G 2021-03-09 Answered
How to solve for third order differential equation of y"7y+6y=2sin(t) using Method of Laplace Transform when y(0)=0,y(0)=0,y"(0)=0?
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lamanocornudaW
Answered 2021-03-10 Author has 85 answers

Step 1
Given third order differential equation is: y"7y+6y=2sin(t)
Apply Laplace transform on each term on both sides of the differential equation.
L{y}7L{y}+6L{y}=2L{sint}
Use the standard appropriate Laplace transforms for the higher order differentials:
L{y}=s3Y(s)s2y(0)sy(0)y(0)
L{y}=sY(s)y(0)
L{y}=Y(s) The Laplace transform of sin(t) is: L{sint}=1(1+s2)
Substitute the formula in the given equation:
[s3Y(s)s2y(0)sy(0)y(0)]7[sY(s)y(0)]+6[Y(s)]=2(1+s2)
Now, substitute the given boundary conditions in the above equation.
(s37s+6)Y(s)=2(1+s2)
Y(s)=2(1+s2)(s37s+6)
Here, the denominator on the right hand side is factorized into:
Y(s)=2(1+s2)(s1)(s2)(s+3)
Step 2 Now, find the inverse Laplace function of Y(s) to get the solution of the differential equation in the form of y(t)
Y(s)=2(1+s2)(s1)(s2)(s+3)
For this equation evaluate the partial fractions:
2(1+s2)(s1)(s2)(s+3)=A(s1)+B(s2)+C(s+3)+D(1+s2)
2=A(1+s2)(s2)(s+3)+B(1+s2)(s1)(s+3)+C(1+s2)(s1)(s2)+D(s1)(s2)(s+3)
For s=1,value of A is: A=15
For s=2,value of B is: B=225
For s=3,value of C is: C=1100
Equating constant terms o both sides gives D value as: D=17100
Then, the Laplace equation turns out to be:
Y(s)=17100(11+s2)15(1s1)+225(1s2)+1100(1s+3)
Now, apply inverse Laplace on each term of the equation.

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