Question

Find y as a function of t if 49y" + 42y' + 13y = 0, y(0) = 8, y'(0) = 5. y(t) =

Answer 1

Answer:

$y(t)=e^{.428 t}(8cos\0.285 t+29.55sin\0.285t)$

Step-by-step explanation:

Given:

  49y''+42y'+13y=0    ,y(0)=8,y'(0)=5  

Lets take  a y''+by'+c=0  is a differential equation.

So auxiliary equation will be

  $a m^2+b m+c=0$  

So according to given problem our  auxiliary equation will be

    $49 m^2+42 m+13=0$  

Then the roots of above equation

$m=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

  But D in the above question is negative so the roots of equation will be imaginary ($D=b^2-4ac$).

By solving m= -0.428+0.285i  , -0.428-0.285i,

  m=$\alpha \pm \beta$

So $y(t)=e^{\alpha t}(C_1cos\beta t+C_2sin\beta t)$

$y(t)=e^{-0.428 t}(C_1cos\0.285 t+C_2sin\0.285t)$

So now by using giving condition we will find

$C_1=8 ,C_2= 29.55$

So  $y(t)=e^{-0.428 t}(8cos\0.285 t+29.55sin\0.285t)$