3.1.11 find the solution of the given initial value problem:

• MHB
• karush
Multiply the second equation by 2 to get \frac{2C_1}{3}+ C_2= 0. Subtract that from the first equation, C_1+ C_2- (\frac{2C_1}{3}+ C_2)= \frac{C_1}{3}= 4. Then C_1= 12. Put that into the original equation, C_1+ C_2= 12+ C_2= 4 so C_2= -8. The specific solution to the differential equation satisfying the initial conditions is y(x)= 12e^{x/3}- 8e^{x/2}.
karush
Gold Member
MHB
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)

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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?

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

karush said:
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

karush said:
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 $$C_2$$.

1. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function that satisfies both the equation and a set of initial conditions. These initial conditions typically include a specific value for the function at a given point, as well as information about the derivative of the function at that point.

2. What does it mean to "find the solution" of an initial value problem?

Finding the solution of an initial value problem means finding a function that satisfies both the equation and the given initial conditions. This function should be able to accurately model the behavior of the system described by the differential equation.

3. How do you solve an initial value problem?

There are various methods for solving initial value problems, including separation of variables, substitution, and using integrating factors. The specific method used will depend on the form of the equation and the initial conditions given.

4. What is the importance of initial value problems in science?

Initial value problems are important in science because they allow us to model and understand the behavior of complex systems. Many physical phenomena, such as population growth, chemical reactions, and electrical circuits, can be described using differential equations and solved using initial value problems.

5. Can initial value problems have multiple solutions?

Yes, it is possible for initial value problems to have multiple solutions. This can occur when the equation is non-linear or when the initial conditions are not unique. In these cases, it is important to carefully consider the given conditions and the behavior of the system to determine the most appropriate solution.

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