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

#### HallsofIvy

##### Well-known member
MHB Math Helper
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

##### Well-known member
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
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 $$\displaystyle 4 = C_1 + C_2$$

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
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 $$C_2$$.