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Sagot :
The general soultion of the given differential equation 12y'' − 5y' − 3y = 0 is
y = [tex]c_{1}e^{\frac{-1}{3}x } +c_{2}e^{\frac{3}{4} x}[/tex].
According to the given question.
We have a differential equation
[tex]12y^{"} -5y^{'} -3y = 0[/tex]
The above equation can be written as
[tex](12D^{2} -5D -3)y = 0[/tex]
The auxillary equation for the above differential equation is
[tex]12m^{2} -5m-3 = 0[/tex]
[tex]\implies 12m^{2} -9m+4m -3 = 0[/tex]
⇒ 3m(4m - 3) +1(4m - 3) = 0
⇒ (3m + 1)(4m -3) = 0
⇒ m = -1/3 or 3/4
Therefore,
[tex]C.f = c_{1}e^{\frac{-1}{3}x } +c_{2}e^{\frac{3}{4} x}[/tex]
Here, P.I = 0
As we know that the general solution of the differential equation is given by
y = PI + CF
Thereofre, the general soultion of the given differential equation 12y'' − 5y' − 3y = 0 is
y = [tex]c_{1}e^{\frac{-1}{3}x } +c_{2}e^{\frac{3}{4} x}[/tex].
Find out more information about general solution of the differential equation here:
https://brainly.com/question/4537000
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