Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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
#SPJ4
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
