# Solving 2nd order DE with initial condition

In summary, the conversation covers two problems involving second order differential equations and finding the values of k that satisfy the given equations. The first problem involves finding the general solution for a specific value of k, while the second problem involves finding multiple values of k for which the given function is a solution to the differential equation. The conversation also discusses the use of an exponential solution and the process of isolating k to find the values of β that satisfy the differential equation.
Hello Guys, We haven't yet covered on how to solve 2nd order equation in class however we have this assignment given to us. Any tips would be appreciated for these 2 little problems.

1. Homework Statement

We have this initial Equation: d2y/dt2−7dy/dt+ky=0, and we need to find the values of k in which the solution y=e3t applies and the general solution.

## The Attempt at a Solution

[/B]
In this case, I simply started to find k but substituting y into the equations.

y(t) = e3t, y'(t) = 3*e3t, y''(t) = 9*e3t

We get: 9e3t - 7*(3*e3t) + k*e3t = 0

=> 9*e3t-21*e3t + k*e3t = 0
=>e3t * (k-12) = 0.

I find the value of K in which y = e3t is a solution to be 12.

Where I am lost is to find the general solution? Do I already have the necessary information?

## Homework Statement

[/B]
Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+11y=0. Hint: There are more than 2 values of k

## The Attempt at a Solution

We know that y is a solution of the DE.

y'(t) = sin(kt), y'(t) = kcos(kt), y''(t)= -k2sin(kt)

By substitution:

-k2sin(kt) + 11sin(kt) = 0
=> sin(kt)(11-k2)=0

I find that √11 for k is a solution, 0 is a solution, and any multiple of π is a solution as well. I tried entering the following:
√11, 0, nπ

And It is not correct. Any tips to see if I did something wrong?Thanks for your help!

Where I am lost is to find the general solution? Do I already have the necessary information?
You are given a hint that the solution involves an exponential. What if you tried a solution that looks like y = eβt and were asked to find what β is? The solutions (there are two of them because this is a second order diff. eq.) should be a function of k.
For the other problem, why have you considered only positive values for k?

kuruman said:
For the other problem, why have you considered only positive values for k?

doesn't n*pi/t cover that? we need sin(kt) = 0 so we need kt = 0 or any multiple of pi right? so let's assume kt= n*pi for n part of ℤ. therefore k = n*pi/t for any n in ℤ

EDIT: Oooohhh... do you mean + or - sqrt(11)?

For the First problem, Let's see.. I got y=eβt, y'=βeβt, y''=β2eβt

By substitution we get eβt2-7β+k) = 0

The only way this is true is that (β2-7β+k) = 0 and if we Isolate k we get: k = β(7-β) But how does that bring me closer to getting a general solution y = something ?

2-7β+k) = 0 and if we Isolate k we get: k = β(7-β)
You are not looking for k, it is assumed given and part of the general solution. The other part are the values of β that satisfy the diff. eq. For what values of β is the (quadratic) equation β2-7β+k = 0 satisfied?

DIT: Oooohhh... do you mean + or - sqrt(11)?
Yes.

kuruman said:
You are not looking for k, it is assumed given and part of the general solution. The other part are the values of β that satisfy the diff. eq. For what values of β is the (quadratic) equation β2-7β+k = 0 satisfied?

Mmh I get β = (7 ± √(49-4k))/2 I am not seeing how you could simplify it more. Are you suggesting that I replace the β in the solution y = eβt?

Yes, that's what I am suggesting. Note that there two solutions, one with a plus sign in front of the radical and one with a minus sign. Also note that when k = 12, the radical is 1 in which case β1 = 3 (you knew that already) and β2 = 4 (new solution).

Thanks, I understand now!

Good. If you haven't formally solved differential equations in class, this problem is a good introduction that will help you make sense of what follows.

## 1. How do I solve a 2nd order differential equation with initial conditions?

To solve a 2nd order differential equation with initial conditions, you can use the method of undetermined coefficients or variation of parameters. Both methods involve finding the complementary function and particular integral, and then adding them together to get the general solution. Finally, you can plug in the initial conditions to find the specific solution.

## 2. What is the difference between a homogeneous and non-homogeneous 2nd order differential equation?

A homogeneous 2nd order differential equation is one where all terms contain only the dependent variable and its derivatives, while a non-homogeneous equation has additional terms that may include constants or other functions. The solution to a homogeneous equation will always be in the form of an exponential function, while the solution to a non-homogeneous equation will also include a particular integral.

## 3. Can I use the Laplace transform to solve a 2nd order differential equation with initial conditions?

Yes, the Laplace transform can be used to solve a 2nd order differential equation with initial conditions. However, you may need to use an additional property, the initial value theorem, to find the specific solution after finding the general solution using the Laplace transform.

## 4. How do I handle complex roots when solving a 2nd order differential equation with initial conditions?

If the characteristic equation of the differential equation has complex roots, you can use Euler's formula to simplify the solution. This involves writing the complex roots in terms of sine and cosine functions. You can then use the initial conditions to find the values of the constants in the solution.

## 5. Is there a specific order in which I should solve a 2nd order differential equation with initial conditions?

The order in which you solve a 2nd order differential equation with initial conditions does not matter. However, it is important to follow the correct steps and methods, such as finding the complementary function and particular integral, and plugging in the initial conditions to find the specific solution. It is also important to check your work and make sure your solution satisfies the original differential equation and initial conditions.

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