Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(O) = 2.8 y(t) =( Preview

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asked Mar 4, 2020 3.1k views

1 Answer

Jan Goyvaerts · Answer 1 · 2020-03-09T20:23:22+0000

Answer:

$y=(4)/(5)e^(t)-(4)/(5)e^{-(2)/(5)t}$

Explanation:

The given equation 5y'' + 3y' - 2y =0 can be written as

(5D^(2)+3D-2)y(t)=0

Solving for complementary function we have Roots of
(5D^(2)+3D-2) as follows

(5D^(2)+5D-2D-2)

$5D(D+1)-2(D+1)=0\\\\(5D-2)(D+1)=0\\\\\therefore D=-1\\D=+2/5$

Thus the complementary function becomes

y=
$y=c_(1)e^{m_(1)t}+c_(2)e^{m_(2)t}$

where

m_(1),m_(2) are calculated roots

thus solution becomes

$y=c_(1)e^(-t)+c_(2)e^{(2)/(5)t}$

Now to solve for the coefficients we use the given boundary conditions

$y(0)=0\\\\\therefore c_(1)+c_(2)=0\\\\y'(0)=-c_(1)+(2)/(5)c_(2)=2.8\\\\\therefore c_(2)+(2)/(5)c_(2)=2.8\\\\c_(2)=2\\\\\therefore c_(1)=-2}$

hence the solution becomes

$y=-2e^-{t}+2e^{(2)/(5)t}$