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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

- Thread starter karush
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- #1

- Jan 31, 2012

- 2,886

- Jan 29, 2012

- 1,151

For one thing, a differential equation is not

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?

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- Jan 31, 2012

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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

- Aug 30, 2012

- 1,180

Pretty much plug-n-chug.

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

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

- Jan 29, 2012

- 1,151

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].why do you have y(x)??

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].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

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].