Initial Value Problem (complex example)

In summary: It looks like you want to use separation of variables here. Do you know how to get all y and dy to one side of the equation, and all x and dx to the other?If you separate the variables, everything should be smooth sailing after that. :smile:
  • #1
King_Silver
83
6
I know the method and can solve other initial value problems. This is the question given:
dy/dx + y(-2) Sin(3x) = 0 for t > 0, with y(0) = 2.

I've brought the dy/dx and let it equal to the rest of the expression so it is now:
dy/dx = -y-2 Sin(3x) , with y(0) = 2 (i.e. when x = 0, y = 2 )

The -y-2 is obstructing me from integrating both sides and solving for constant C using the information given. What would I have to do to eliminate the -y-2 entirely from the question so I can continue standard procedures?
 
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  • #2
Maybe I'm overlooking something, but can't you just multiply the original equation by ##y^2##?
 
  • #3
Krylov said:
Maybe I'm overlooking something, but can't you just multiply the original equation by ##y^2##?
But if you multiply the original by y2 then you will need to do the same to the left hand side as you have done to the right hand side, meaning the dy/dx would also be affected.
 
  • #4
King_Silver said:
I know the method and can solve other initial value problems. This is the question given:
dy/dx + y(-2) Sin(3x) = 0 for t > 0, with y(0) = 2.

It's not clear which is the dependent variable here. The equation suggests x, but there's this 't > 0' restriction. I'm confused.
I've brought the dy/dx and let it equal to the rest of the expression so it is now:
dy/dx = -y-2 Sin(3x) , with y(0) = 2 (i.e. when x = 0, y = 2 )

It looks like you want to use separation of variables here. Do you know how to get all y and dy to one side of the equation, and all x and dx to the other?
The -y-2 is obstructing me from integrating both sides and solving for constant C using the information given. What would I have to do to eliminate the -y-2 entirely from the question so I can continue standard procedures?

If you separate the variables, everything should be smooth sailing after that. :smile:
 
  • #5
SteamKing said:
It's not clear which is the dependent variable here. The equation suggests x, but there's this 't > 0' restriction. I'm confused.It looks like you want to use separation of variables here. Do you know how to get all y and dy to one side of the equation, and all x and dx to the other?If you separate the variables, everything should be smooth sailing after that. :smile:

What do you mean by that? all y to the left and all x terms to the right?

so like ∫ y-2 dy = ∫Sin(3x) dx ? :)
 
  • #6
SteamKing said:
If you separate the variables, everything should be smooth sailing after that. :smile:
Indeed. Just allow yourself to do the usual abuse of differentials.
 
  • #7
King_Silver said:
What do you mean by that? all y to the left and all x terms to the right?

so like ∫ y-2 ∫dy = Sin(3x) dx ? :)
I think you have too many integral signs on one side of the equation, and not enough on t'other.
 
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  • #8
SteamKing said:
I think you have too many integral signs on one side of the equation, and not enough on t'other.
Sorry that was just an accident edited the post :) thanks! got it sorted now cheers! :D
 
  • #9
King_Silver said:
What do you mean by that? all y to the left and all x terms to the right?

so like ∫ y-2 dy = ∫Sin(3x) dx ? :)
Shouldn't that be ∫ y+2 dy?
 
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  • #10
Samy_A said:
Shouldn't that be ∫ y+2 dy?
And a -Sin(3x) I think, sorry my bad!
I'm sure it should be a -Sin(3x) as well.
 
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  • #11
integrated it.
y3/3 = 1/3 Cos(3x) + C, with y(0) = 2

so 23/3 = 1/3 Cos(0) + C
2 2/3 = 1/3 + C

C = 2 1/3

That is what I got :)
 

FAQ: Initial Value Problem (complex example)

1. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function or set of functions that satisfy certain initial conditions. These conditions are typically values of the function(s) and their derivatives at a specific point or interval.

2. Can you give an example of an initial value problem?

One example of an initial value problem is the classic "population growth" problem, where the rate of change of a population is proportional to the size of the population. The initial value problem would involve finding the population function given the initial population size and growth rate.

3. What makes an initial value problem complex?

An initial value problem can become complex when the differential equation involved is nonlinear or involves higher order derivatives. Additionally, complex initial conditions or boundary conditions can also make an initial value problem more difficult to solve.

4. What methods can be used to solve initial value problems?

There are several methods that can be used to solve initial value problems, including analytical methods such as separation of variables and numerical methods such as Euler's method or the Runge-Kutta method. The choice of method depends on the complexity of the problem and the desired level of accuracy.

5. How are initial value problems used in real-world applications?

Initial value problems have many applications in fields such as physics, engineering, and economics. They can be used to model and predict various phenomena, from population growth to heat transfer to electrical circuits. By finding solutions to initial value problems, scientists and engineers can make predictions and optimize designs in these fields.

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