Question

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

Answer 1

Answer:

$y=\frac{4}{5}e^{t}-\frac{4}{5}e^{-\frac{2}{5}t}$

Step-by-step 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

$y=c_{1}e^{-t}+c_{2}e^{\frac{2}{5}t}$

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

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

hence the solution becomes

$y=-2e^-{t}+2e^{\frac{2}{5}t}$