Substitution in Differential Equations

In summary, the conversation discusses finding a solution to the differential equation dy/dx=sin(x-y). Through appropriate substitutions and integration, the solution is found to be x=tan(x-y) + sec(x-y). The conversation also discusses discrepancies between the solution found and the one given in the book.
  • #1
rgalvan2
30
0
Make an appropriate substitution to find a solution of the equation dy/dx=sin(x-y). Does this general solution contain the linear solution y(x)=x-pi/2 that is readily verified by substitution in the differential equation?

Here's what I did:
v=x-y
y=x-v
y'=1-dv/dx

1-dv/dx=sin(v)
1-sin(v)=dv/dx
dx=dv/(1-sin(v))
x=2/(cot(v/2)-1)

The solution in the back of the book gives:
x=tan(x-y) + sec(x-y)

What am I doing wrong?
Any help is greatly appreciated.
Thanks!
 
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  • #2
f =

1/(1-sin(x))


>> int(f,x)

ans =

-2/(tan(1/2*x)-1)

I have something different from MATLAB when I integrate 1/(1-sin(x))
 
  • #3
rgalvan2 said:
Make an appropriate substitution to find a solution of the equation dy/dx=sin(x-y). Does this general solution contain the linear solution y(x)=x-pi/2 that is readily verified by substitution in the differential equation?

Here's what I did:
v=x-y
y=x-v
y'=1-dv/dx

1-dv/dx=sin(v)
1-sin(v)=dv/dx
dx=dv/(1-sin(v))
Everything is fine to here, but goes downhill after that.
The integral on the right isn't too bad.

[tex]\int \frac{dv}{1 - sin(v)}[/tex]
[tex]= \int \frac{dv}{1 - sin(v)} * \frac{1 + sin(v)}{1 + sin(v)}[/tex]
[tex]=\int \frac{(1 + sin(v))dv}{1 - sin^2(v)}[/tex]

The denominator simplifies to cos^2(v) and you can split the integral into two integrals, one of which is straightforward. The other one requires only an ordinary substitution.

I ended with the same answer as in the book.

rgalvan2 said:
x=2/(cot(v/2)-1)

The solution in the back of the book gives:
x=tan(x-y) + sec(x-y)

What am I doing wrong?
Any help is greatly appreciated.
Thanks!
 

Related to Substitution in Differential Equations

1. What is substitution in differential equations?

Substitution in differential equations is a method used to solve differential equations by replacing the original variable with a new variable. This allows us to simplify the equation and make it easier to solve.

2. Why is substitution used in solving differential equations?

Substitution is used in solving differential equations because it helps to reduce the complexity of the equation and make it easier to solve. It also allows us to transform the equation into a more manageable form, which can lead to a more straightforward solution.

3. How does substitution work in solving differential equations?

Substitution works by replacing the original variable in the differential equation with a new variable. This new variable is chosen based on its ability to simplify the equation and make it easier to solve. After substitution, the equation is then solved using traditional methods such as separation of variables or integrating factors.

4. What are the benefits of using substitution in solving differential equations?

The main benefit of using substitution in solving differential equations is that it simplifies the equation and makes it easier to solve. This can save time and effort when trying to find a solution. Additionally, substitution can also help to reveal patterns and relationships within the equation that may not have been apparent before.

5. Are there any limitations to using substitution in solving differential equations?

Yes, there are some limitations to using substitution in solving differential equations. One limitation is that it may not work for all types of differential equations. Additionally, the choice of the new variable can also greatly impact the ease of solving the equation. In some cases, substitution may also lead to more complex equations that are more difficult to solve.

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