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प्रश्न
Solve the following differential equation:
`(4"D"^2 + 16"D" + 15)y = 4"e"^((-3)/2x)`
उत्तर
The auxiliary equation is 4m2 + 16m + 15 = 0
4m2 + 16m + 10m + 15 = 0
2m(2m + 3) + 5(2m + 3) = 0
(2m + 3)(2m + 5) = 0
2m = – 3, – 5
∴ m = `(-3)/2, (-5)/2`
Roots are real and different
C.F = (Ax + B) em1x + Bem2x
C.F = `"Ae"^((-3)/2 x) + "Be"^((-5)/2 x)`
P.I = `1/((4"D"^2 + 16"D" + 15)) 4"e"^((-3)/2 x)`
Replace D by `(-3)/2`
4D2 + 16D + 15 = 0
When D = `(-3)/2`
P.I = `x * 4/(4(2"D") + 16(1)) 4"e"^((-3)/2x)`
= `x * 1/(8"D" + 16) 4"e"^((-3)/2x)`
= `x * 1/(8((-3)/2) + 16) 4"e"^((-3)/2 x)`
= `x * 1/(-12 + 16) 4"e"^((-3)/2 x)`
= `x * 1/4 4"e"^((-3)/2 x)`
P.I = `x"e"^((-3)/2 x)`
The general solution is y = C.F + P.I
y = `"Ae"^((-3)/2 x) + "Be"^((-5)/2 x) + x"e"^((-3)/2 x)`
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