Bobosharif S. answered • 04/28/18
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Mathematics/Statistics Tutor
y''- y' - 2y = 5sin(x) ; y(0) = 1 , y'(0) = -1
First we find the general solution of the Homogenous Equation (HE)
y''- y' - 2y = 0
The characteristic equation is
λ2-λ-2=0
λ1=-1,λ2=2
So the general solution of HE is
yg(x)=C1e-x+C2e2x
Now e need to find a particular solution of nohmogenonous equation
y''- y' - 2y = 5sin(x)
We searh the solutin yj the form y=Asin(x)+BCos(x). Sunstitutiion into the equation gives A=1/2,B=-3/2 and the geberal solution would be
y(x)=C1e-x +C2e2x+(1/2) (cos(x)-3sin(x))
Now by using initial conditions, find C1 and C2.
Bobosharif S.
Laplace Transform (LP)
You have to use properties of LP:
L{y'}=sL{y(s)}-y(0)
L{y''}=s2L{y(s)}-sy(0)-y'(0)
L{5sin(x)}=5/(1+s2)
Take LP from both sides of
y''- y' - 2y = 5sin(x)
L{y''}-L{y'}-2L{y}=5L{sin(x)}
s2L{y}-sy(0)-y'(0)-sL{y}+sy(0)-2L{y}=5/(1+s2)
L{y}(s2-s-2)=5/(1+s2)-1
L{y}=(4-s2)/(1+s2) (1/(s2-s-2))
L{y}=-(2+s)/(1+s+s2+s3)
So far what we did is LP from the solution of DE. To find solution intself, we haveto take inverse LP:
y(x)=L-1{-(2+s)/(1+s+s2+s3)}
y(x)=-e-x/2+(1/2)(cos(x)-3sin(x))
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04/29/18
Phantom L.
Thanks Mr. Sharif
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04/29/18
Phantom L.
04/28/18