Find the particular solution of a non-homogeneous equation using Laplace transformation, y(0) = 0, y'(0)=0, y"(0)=0 y'''-5y''-22y'+56y=12-32e^{-8x}+2e^{4x}

Emeli Hagan 2020-12-07 Answered
Find the particular solution of a non-homogeneous equation using Laplace transformation,
y(0)=0,
y(0)=0,
y"(0)=0
y5y22y+56y=1232e8x+2e4x
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Expert Answer

unett
Answered 2020-12-08 Author has 119 answers
Step 1
Given non homogenous equation
y5y22y+56y=1232e8x+2e4x
With initial conditions y(0)=0,y(0)=0,y"(0)=0
Taking Laplace transform on both sides,
L{y}5L{y}22L{y}+56L{y}=L{12}32L{e8x}+2L{e4x}
s3y¯5s2y¯22sy¯+56y¯=12s32(s+8)+2(s4)
y¯(s35s222s+56)=12s32(s+8)+2(s4)
y¯(s2)(s+4)(s7)=12s32(s+8)+2(s4)
y¯=12s(s2)(s+4)(s7)32(s+8)(s2)(s+4)(s7)+2(s4)(s2)(s+4)(s7)
Dividing each term on right into partial fraction,
y¯=(314s15(s2)122(s+4)+12385(s7))(475(s+8)875(s2)+133(s+4)+32825(s7))+(124(s4)+130(s2)1264(s+4)+2165(s7))
Taking inverse Laplace transform on both sides,
y(x)=31415e2x122e4x+12385e7x+475e8x+875e2x433e4x32825e7x124e4x+130e2x1264e4x+2165e7x

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