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Question
Solve `(d^2y)/dx^2-16y=x^2 e^(3x)+e^(2x)-cos3x+2^x`
Solution
The auxiliary equation is 𝐷2 − 16 = 0
∴` D=4,-4`
∴ The C.F.is `y= C_1e ^(4x)+C_2 e^(-4x)`
Now, to find P.I.,
`P.I.= 1/(D^2-16)(x^2 e^(3x)+e^(02x)-cos 3x+2^x)`
Now,` 1/(D^2-16) x^2 e^(3x)=e^(3x).1/((D+3)^2-16).x^2`
= `e^3x. 1/(D^2+6D+9-16).x^2=e^(3x). 1/(D^2+6D-7).x^2`
=`-e^(3x)/7. 1/((1D^2+6D)/7).x^2`
= `- e^(3x)/7. (1-(D^2+6D)/7)^-1.x^2`
=`-e^(3x)/7 (1+(D^2+6D)/7+(D^4+6D^3+36D^2)/49+...).x^2`
=`-e^(3x)/7(x^2+(12x+2)/7+72/49)=-e^(3x)/7(x^2+(12x)/7+86/49)`
∴ `1/(D^2-16).e^(2x)=e^(2x) 1/(2^2-16)=e^(2x).1/(2^2-16)=e^(2x).1/12`
∴ `1/(D^2-16).cos 3x=(cos 3x)/(-9-16)=(cos3x)/-25`
∴ `1/(D^2-16).2^x=1/(D^2-16).e^(x log 2)=e^(x log2)/((log 2)^2-16).=2^x/((log2)^2-16)`
`P.I.=-e^(3x)/7(x^2+(12x)/7+86/49)+e^(2x).1/12+(cos 3x)/25+2^x/((log 2)^2-16)`
∴ The complete equation is,
`y=C_1e^(4x)+C_2e^(-4x)-e^(3x)/7(x^2+(12x)/7+86/49)+e^(2x).1/12+(cos3x)/25+2^x/((log 2)^2-16)`
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