Question

Find y as a function of t if y" + 14y' + 85y = 0, y(0) = 9, y'(0) = 8. y(t) =

Answer 1

Answer:

$y=9e^{-7x}cos(6x)+\frac{71}{6}e^{-7x}sin(6x)$

Step-by-step explanation:

The Given diffrential equation can be written as follows

$(D^{2}+14D+85)y=0$

Solving the auxiliary equation we get

$D^{2}+14D+85=0$

Comparing with the standard quadratic equation $ax^{2}+bx+c=0$ we get

$D=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Using the values we get the roots as $-7\pm 6i$

The roots are imaginary and conjugate

For a auxillary equation with roots $a\pm bi$ the solution is given by

$y=c_{1}e^{ax}cos(bx)+c_{2}e^{ax}sin(bx)$

Applying values we get the solution of the equation as

$y=c_{1}e^{-7x}cos(6x)+c_{2}e^{-7x}sin(6x)$....................(i)

here $c_{1},c_{2)$ are constants whose values shall be obtained using the given boundary conditions y(0) = 9 ; y'(0) = 8

In equation i putting x = 0 we get

9 =$c_{1}$

Now we have

$y'=-7c_{1}e^{-7x}cos(6x)-6c_{1}e^{-7x}sin(6x)-7c_{2}e^{-7x}sin(6x)+6c_{2}e^{-7x}cos(6x)$

Now y'(0) = 8

Thus we get

$8=-7c_{1}+6c_{2}$

Solving for $c_{2}$ we get

$c_{2}=71/6$

Thus the solution becomes

$y=9e^{-7x}cos(6x)+\frac{71}{6}e^{-7x}sin(6x)$