Solve the differential equation ..(b and a are variables) (D^2+4D+3)b=1+2a+3a^2

Cabiolab

Cabiolab

Answered question

2021-09-11

Solve the differential equation ..(b and a are variables)
(D2+4D+3)b=1+2a+3a2

Answer & Explanation

Viktor Wiley

Viktor Wiley

Skilled2021-09-12Added 84 answers

Step 1
Given differential equation
(D2+4D+3)b=1+2a+3a2  where  D=dda
The auxillicory equation is given by
m2+4m+3=0
m2+3m+m+3=0
m(m+3)+1(m+3)=0
(m+3)(m+1)=0
m1=3  and  m2=1
Hence the complementary function is given by
yc=c1em1a+c2em2a
yc=c1e3a+c2ea
Now we find the Particular intesteed
yp=1F(D)f(a)
=1D2+4D+3(1+2a+3a2)
=13(1+D2+4D3)(1+2a+3a2)
=13(1D2+4D3+D2+4D3)2s˙)(1+2a+3a2)
=13[1(D23+42D3)]+D4+16D2+8D39s˙](1+2a+3a2)

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