Question

Find the real-valued solution to the initial value problem

7y''+16y'+16y=0 y(0)=0 y'(0)=5

Answer 1

First Solve
\[
7 r^2 + 16 r + 16 = 0\\
\begin{array}{c}
r=\frac{4}{7} \left(-2-i
\sqrt{3}\right) \\
r=\frac{4}{7} \left(-2+i
\sqrt{3}\right) \\
\end{array}


\]
Can you finish it from here?

Answer 2

First, thank you for replying! and Second, I got up to that and plugged it into the equation and got stuck

Answer 3

I ended up with y(x)=e^(-8/7x)[C1cos((8/14x(sqrt3))+C2sin((8/14)x(sqrt(3))

Answer 4

This the general solution
\[
y=A e^{-8 x/7} \cos
\left(\frac{4 \sqrt{3}
x}{7}\right)+B e^{-8 x/7}
\sin \left(\frac{4 \sqrt{3}
x}{7}\right)
\]
Find A and B from the initial condition.

Answer 5

\[
0 = y(0)= A
\]
So A =0
Try to get B alone. I will check when I come back form shopping.

Answer 6

ok thanks so much!

Answer 7

i solved for y' and ended up getting B=-32.65544457

Answer 8

and it's sadly incorrect

Answer 9

\[
y'=-\frac{4}{7} \sqrt{3} A e^{-8 x/7} \sin
\left(\frac{4 \sqrt{3} x}{7}\right)-\frac{8}{7}
A e^{-8 x/7} \cos \left(\frac{4 \sqrt{3}
x}{7}\right)-\frac{8}{7} B e^{-8 x/7} \sin
\left(\frac{4 \sqrt{3} x}{7}\right)+\frac{4}{7}
\sqrt{3} B e^{-8 x/7} \cos \left(\frac{4
\sqrt{3} x}{7}\right)\\

y'(0)=\frac{4 \sqrt{3} B}{7}-\frac{8 A}{7}=\frac{4 \sqrt{3} B}{7}=5\\


B=\frac{35}{4 \sqrt{3}}

\]

Answer 10

\[
y=\frac{35 e^{-8 x/7} \sin \left(\frac{4 \sqrt{3}
x}{7}\right)}{4 \sqrt{3}}
\]
is your final answer.

Answer 11

thank you!