Calculus II - Solving Second Order Differential Equation

In summary, none of the given functions, a) y=cos(x), b) y=sin(x), c) y=1/2*xcos(x), d) y=1/2*x*sin(x), are solutions to the differential equation y''+y=sin(x). This type of equation is known as an inhomogeneous second order ordinary differential equation with constant coefficients, and is not a standard topic in Calculus II. It is typically taught in first-year undergraduate math.
  • #1
GreenPrint
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Homework Statement



Which of the following functions are soltuions of the differntial equation y''+y=sin(x)?
a) y=cos(x) b) y=sin(x) c) y=1/2*xcos(x) d) y=1/2*x*sin(x)

Homework Equations





The Attempt at a Solution



I'm kind of lost on how to solve this problem. I don't think this is a standard calculus II problem but it was on a packet that we were asked to solve.

I learned from the internet that you can solve a second differential order equation of form
Af''(x) + Bf'(x) + Cf(x) = 0
By setting f'(x) as r_1, f''(x) as r_2, f(x) as 1
then
Ar^2 + Br + C = 0
and plugging into
y(x) = c_1 e^(r_1*x) + c_2*e^(r_2*x)
were c_1 and c_2 are just some constants that can be solved for if your given initial conditions

I'm unsure how to solve this problem because it's of a different form of
f''(x) + f(x) = sin(x)
I'm unsure what to do about the sin(x) term in this case and not sure what to do sense I don't have initial condition and how to come up with the constant terms as a result...

Thanks for any Help!
 
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  • #3
Thanks for your response. No I have never seen type of equation before. Is it a standard calculus II topic? That pdf file looks very useful, I'll see if I can solve the problem after looking over it.
 
  • #4
GreenPrint said:
Thanks for your response. No I have never seen type of equation before. Is it a standard calculus II topic? That pdf file looks very useful, I'll see if I can solve the problem after looking over it.
I am not familiar with the US education system, so I can't comment. However, over here in the UK, this is first year undergraduate level.
 
  • #5
GreenPrint said:

Homework Statement



Which of the following functions are soltuions of the differential equation y''+y=sin(x)?
a) y=cos(x) b) y=sin(x) c) y=1/2*xcos(x) d) y=1/2*x*sin(x)

Thanks for any Help!

You can also, when checking for solutions, think in the other direction, which makes the calculation quite simple.

find y'' of the suggested y=...

Calculate y''+y and see if this equals sin(x)

I think you'll find that's far easier than solving the 2nd order differential equation. :cool:

When I learned about diff eqs this spring we started approaching possible solutions this way, later we learn't how to solve such differential equations.
 
  • #6
You are not being asked to solve the differential equation (inhomogeneous second-order ordinary differential equations are not a Calculus II topic), but to "verify" a solution. Which choice works in the equation?
 
  • #7
I'm actually getting none of them

a) y=cos(x)
y'=-sin(x)
y''=-cos(x)
y''+y=-cos(x)+cos(x)=0

b)y=sin(x)
y'=cos(x)
y''=-sin(x)
y''+y=-sin(x)+sin(x)=0

c)y=(xcos(x))/2
y'=(-xsin(x))/2+cos(x)/2
y''=-(xcos(x))/2-sin(x)
y''+y=(-xcos(x))/2-sin(x)+(xcos(x))/2=-sin(x)

d)y=(xsin(x))/2
y'=sin(x)/2+(xcos(x))/2
y''=cos(x)/2+cos(x)/2-(xsin(x))/2=cos(x)-(xsin(x))/2
y''+y=cos(x)-(xsin(x))/2+(xsin(x))/=cos(x)

Is the problem missed up or am I doing something wrong because I'm getting that all the answers are wrong?
 
  • #8
I can't find something wrong there...
 
  • #9
As you show the problem, none of the given functions is a solution. It would be a good idea to check that you wrote them down correctly, particularly the ones for parts c and d. If you missed a sign, that would affect your answer.
 
  • #10
dynamicsolo said:
You are not being asked to solve the differential equation (inhomogeneous second-order ordinary differential equations are not a Calculus II topic), but to "verify" a solution. Which choice works in the equation?
None of them do, which is why I suggested that the OP look into how to solve the problem himself. I wasn't sure whether that was part of the exercise or not.
 
  • #11
Hootenanny said:
None of them do, which is why I suggested that the OP look into how to solve the problem himself. I wasn't sure whether that was part of the exercise or not.

If it's a multiple-choice question for practice, finding the error is probably not part of the exercise.

I guess I want to ask whether this set of problems came from a published source or from an instructor. It wouldn't be the first time I'd seen someone make up a bunch of problems and not check for typoes or whether a problem could even be solved as stated. I suspect the answer was intended to be (c), but a minus sign was omitted on the inhomogeneous term or on the solution. [I agree that there are no errors in GreenPrint's calculations.]
 
  • #12
It actually came from my professor, I guess the problem is messed up, I'm glad I didn't do something wrong =). It's strange, the difference between education systems, it's something taught in first year undergraduate math but not in the US apparently... I guess they have better education systems elsewhere lol, that doesn't surprise me
 
  • #13
dynamicsolo said:
If it's a multiple-choice question for practice, finding the error is probably not part of the exercise.

I guess I want to ask whether this set of problems came from a published source or from an instructor. It wouldn't be the first time I'd seen someone make up a bunch of problems and not check for typoes or whether a problem could even be solved as stated. I suspect the answer was intended to be (c), but a minus sign was omitted on the inhomogeneous term or on the solution. [I agree that there are no errors in GreenPrint's calculations.]
Fair point. I just didn't want to give the answer away if this wasn't the case :smile:
 
  • #14
Solving the differential equation is probably too advanced for a Calc II course, but verifying that a given function is a solution of a differential equation is appropriate, IMO. For problems like this, all you're doing is finding out whether the given function and it's relevant derivatives make the differential equation identically true.
 
  • #15
GreenPrint said:
It actually came from my professor, I guess the problem is messed up, I'm glad I didn't do something wrong =). It's strange, the difference between education systems, it's something taught in first year undergraduate math but not in the US apparently... I guess they have better education systems elsewhere lol, that doesn't surprise me
I seem to recall the variation of parameters method was covered in the second semester of calculus when I took it. You can use that method to solve this equation.
 

FAQ: Calculus II - Solving Second Order Differential Equation

1. What is a second order differential equation?

A second order differential equation is a mathematical equation that involves the second derivative of a function. In other words, it is an equation that relates the rate of change of a function to its second derivative.

2. Why is it important to solve second order differential equations?

Second order differential equations are important because they are used to model many physical systems in science and engineering. They also have a wide range of applications in fields such as economics, biology, and chemistry.

3. What methods can be used to solve second order differential equations?

The main methods used to solve second order differential equations are the substitution method, the variation of parameters method, and the method of undetermined coefficients. Other methods may also be used, depending on the specific equation.

4. Can second order differential equations have multiple solutions?

Yes, second order differential equations can have multiple solutions. This is because the general solution of a second order differential equation contains two arbitrary constants, which can take on different values and lead to different solutions.

5. How can I check if my solution to a second order differential equation is correct?

To check if a solution to a second order differential equation is correct, you can substitute it back into the original equation and see if it satisfies the equation. You can also plot the solution and its derivatives to visually confirm if it is a valid solution.

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