MHB 3.1.11 find the solution of the given initial value problem:

[karush]

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

 

[HallsofIvy]

First of all, your fourth line, "6y''- 5y'+ y= (r- 3)(r- 2)= 0" is non-sense!

For one thing, a differential equation is not equal to its characteristic equation.
More importantly, you have the wrong characteristic equation. (r- 3)(r- 2)= r^2- 5r+ 6, not 6r^2- 5r+ 1= 0 which is the correct characteristic equation for 6y''- 5y+ 1= 0.

Now that can be factored as (3r- 1)(2r- 1)= 0 which has roots r= 1/3 and r= 1/2. So the general solution to the differential equation is
y(x)= C_1e^{t/3}+ C_2e^{t/2}.

Now can you get to answer 11?

 

[karush]

why do you have y(x)??

also how is the $y'(0)=0$ to be used

I can see that $3\cdot 4 =12 $ and $2\cdot 4= 8$ and the $12-8=4$ and $e^0=1$

but how is that applied to the next steps

 

[topsquark]

why do you have y(x)??

also how is the $y'(0)=0$ to be used

I can see that $3\cdot 4 =12 $ and $2\cdot 4= 8$ and the $12-8=4$ and $e^0=1$

but how is that applied to the next steps

Pretty much plug-n-chug.

We have that
y = C_1 e^{t/3} + C_2 e^{t/2}

The condition y(0) = 4 implies that [math]4 = C_1 + C_2[/math]

Now, [math]y' = \dfrac{C_1}{3} e^{t/3} + \dfrac{C_2}{2} e^{t/2}[/math]

Thus y'(0) = 0 implies that [math]0 = \dfrac{C_1}{3} + \dfrac{C_2}{2}[/math]

Two equations, two unknowns.

-Dan

 

[HallsofIvy]

why do you have y(x)??

Your original problem was to solve "6y''- 5y'+ y= 0". There is no "independent variable" stated so we are free to choose whatever letter we want. y(x)= C_1e^{x/3}+ C_2e^{x/2} is the same as y(t)= C_1e^{t/3}+ C_2e^{t/2}.

also how is the $y'(0)=0$ to be used

I can see that $3\cdot 4 =12 $ and $2\cdot 4= 8$ and the $12-8=4$ and $e^0=1$but how is that applied to the next steps

You are given the initial conditions y(0)= 4, y'(0)= 0 and you now know that the "general solution" to the differential equation is y(x)= C_1e^{x/3}+ C_2e^{x/2}. So y(0)= C_1e^{0/3}+ C_3e^{0/2}= C_1+ C_2= 4.
Differentiating, y'(x)= (C_1/3)e^{x/3}+ (C_2/2)e^{x/3} so y'(0)= \frac{C_1}{3}+ \frac{C_2}{2}= 0.

You now have two equations, C_1+ C_2= 4 and \frac{C_1}{3}+ \frac{C_2}{2}= 0 to solve for C_1 and C_2.