- #1

karush

Gold Member

MHB

- 3,269

- 5

find the solution of the given initial value problem:

$6y''-5y'+y=0\quad y(0)=4 \quad y'(0)=0$

if $r=e^{5t}$ then

$\displaystyle 6y''-5y'+y=(r-3)(r-2)=0$

then

$y=c_1e^{3t}+c_1e^{2t}=0$

for $y(0)=4$

$y(0)=c_1e^{3(0)}+c_1e^{2(0)}=4$

ok I don't see how the last few steps lead to the book answer (11)

View attachment 9019

$6y''-5y'+y=0\quad y(0)=4 \quad y'(0)=0$

if $r=e^{5t}$ then

$\displaystyle 6y''-5y'+y=(r-3)(r-2)=0$

then

$y=c_1e^{3t}+c_1e^{2t}=0$

for $y(0)=4$

$y(0)=c_1e^{3(0)}+c_1e^{2(0)}=4$

ok I don't see how the last few steps lead to the book answer (11)

View attachment 9019