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Find the general solution of the given second-order differential equation. 12y'' − 5y' − 3y = 0

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].

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