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3.1.11 find the solution of the given initial value problem:

karush

Well-known member
Jan 31, 2012
2,886
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)

3_1_11.PNG
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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. [tex](r- 3)(r- 2)= r^2- 5r+ 6[/tex], not [tex]6r^2- 5r+ 1= 0[/tex] 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
[tex]y(x)= C_1e^{t/3}+ C_2e^{t/2}[/tex].

Now can you get to answer 11?
 

karush

Well-known member
Jan 31, 2012
2,886
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,180
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
[tex]y = C_1 e^{t/3} + C_2 e^{t/2}[/tex]

The condition y(0) = 4 implies that \(\displaystyle 4 = C_1 + C_2\)

Now, \(\displaystyle y' = \dfrac{C_1}{3} e^{t/3} + \dfrac{C_2}{2} e^{t/2}\)

Thus y'(0) = 0 implies that \(\displaystyle 0 = \dfrac{C_1}{3} + \dfrac{C_2}{2}\)

Two equations, two unknowns.

-Dan
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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. [tex]y(x)= C_1e^{x/3}+ C_2e^{x/2}[/tex] is the same as [tex]y(t)= C_1e^{t/3}+ C_2e^{t/2}[/tex].


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 [tex]y(x)= C_1e^{x/3}+ C_2e^{x/2}[/tex]. So [tex]y(0)= C_1e^{0/3}+ C_3e^{0/2}= C_1+ C_2= 4[/tex].
Differentiating, [tex]y'(x)= (C_1/3)e^{x/3}+ (C_2/2)e^{x/3}[/tex] so [tex]y'(0)= \frac{C_1}{3}+ \frac{C_2}{2}= 0[/tex].

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