Particular Sub y_p into DE for r(x)

In summary, the given expression is a particular solution of the differential equation y'' + p(x)y' + q(x)y = r(x) when substituted directly into the equation and using the product rule to differentiate. The Wronskian, W, is used in the expression to find the integral of the product of the two independent solutions, y_1 and y_2, and the integration variable should be changed to avoid confusion.
  • #1
Ted123
446
0

Homework Statement



Show (by substituting it directly into the differential equation) that

[itex]\displaystyle y_p = y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx[/itex]

is a particular solution of [itex]y'' + p(x)y' + q(x)y = r(x)[/itex].

Homework Equations



[itex]W[/itex] is the Wronskian [itex]y_1 y_2^{\prime} - y_2 y_1^{\prime}[/itex]

The Attempt at a Solution



How do I sub [itex]y_p[/itex] into the DE?
 
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  • #2
Hi Ted :smile:

Just calculate [tex]y_p^{\prime\prime}+p(x)y_p^\prime+q(x)y_p[/tex] and check that it equals r(x)...
 
  • #3
micromass said:
Hi Ted :smile:

Just calculate [tex]y_p^{\prime\prime}+p(x)y_p^\prime+q(x)y_p[/tex] and check that it equals r(x)...

I can't see how to differentiate [itex]y_p[/itex] and certainly [itex]y_p^{\prime}[/itex] though.

For instance, how do I find:

[itex]\displaystyle \frac{d}{dx} \left( y_2 \int \frac{ry_1}{W}\;dx \right)[/itex] ?

I know

[itex]\frac{d}{dx} \left( \int \frac{ry_1}{W}\;dx \right) = \frac{ry_1}{W}[/itex]

but [itex]y_2[/itex] is a function of x, and it is multiplied by that integral...
 
  • #4
Ah yes. There, you have to differentiate a product of two functions. So you need to use the product rule:

[tex](fg)^\prime=f^\prime g+fg^\prime[/tex]
 
  • #5
micromass said:
Ah yes. There, you have to differentiate a product of two functions. So you need to use the product rule:

[tex](fg)^\prime=f^\prime g+fg^\prime[/tex]

I get:

[itex]\displaystyle y_p^{\prime} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx[/itex]

[itex]\displaystyle y_p^{\prime\prime} = y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx[/itex]

So:

[itex]\displaystyle y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx + p(x) \left( y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx \right) + q(x) \left( y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx \right)[/itex]

should equal r but does it?
 
  • #6
You didn't exactly follow the product rule did you??

You should have done

[tex]\left(y_2\int\frac{ry_1}{W}\right)^\prime=y_2^\prime\int{\frac{ry_1}{W}}+y_2\frac{ry_1}{W}[/tex]

Now, if you differentiate like that, then what is [tex]y_p[/tex] then?
 
  • #7
micromass said:
You didn't exactly follow the product rule did you??

You should have done

[tex]\left(y_2\int\frac{ry_1}{W}\right)^\prime=y_2^\prime\int{\frac{ry_1}{W}}+y_2\frac{ry_1}{W}[/tex]

Now, if you differentiate like that, then what is [tex]y_p[/tex] then?

I did follow the product rule:

[itex]\displaystyle y_p = y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx[/itex]

so differentiating the first term gives what you have.

Then differentiating the 2nd term [itex]- y_1 \int \frac{ry_2}{W}\;dx[/itex] gives:

[itex]- y_1^{\prime} \int \frac{ry_2}{W}\;dx - \frac{ry_1y_2}{W}[/itex]

so adding the 2 together gives:

[itex]y_p^{\prime} = y_2^\prime\int{\frac{ry_1}{W}\;dx}+\frac{ry_1y_2}{W} - y_1^{\prime} \int \frac{ry_2}{W}\;dx - \frac{ry_1y_2}{W} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx[/itex]
 
  • #8
Ah, yes, you are correct! :smile:
Now, try to show that the expression equals r(x). Try to do something with the definition of y1 and y2. (How are these things defined anyway?)
 
  • #9
micromass said:
Ah, yes, you are correct! :smile:
Now, try to show that the expression equals r(x). Try to do something with the definition of y1 and y2. (How are these things defined anyway?)

[itex]y_1(x)[/itex] and [itex]y_2(x)[/itex] are independent solutions of

[itex]y^{\prime\prime} + p(x) y^{\prime} + q(x) y =0[/itex].
 
  • #10
So you can substitute for [tex]y_1''(x)[/tex] in your equation above.

EDIT: even easier, collect terms which have [tex]\int \frac{ry_2}{W}\;\mathrm{d}x[/tex].
 
  • #11
grey_earl said:
So you can substitute for [tex]y_1''(x)[/tex] in your equation above.

EDIT: even easier, collect terms which have [tex]\int \frac{ry_2}{W}\;\mathrm{d}x[/tex].

I'm not entirely sure what you're saying. We want:

[itex]\int \frac{ry_2}{W}\;dx \left( -y_1^{\prime\prime} - py_1^{\prime} - qy_1 \right) + \int \frac{ry_1}{W}\;dx \left( y_2^{\prime\prime} + py_2^{\prime} + qy_2 \right)[/itex]

to equal [itex]r(x)[/itex].

[itex]y_1[/itex] and [itex]y_2[/itex] are linearly independent solutions of [itex]y^{\prime\prime} + py^{\prime} + qy = r[/itex].

If we substitute [itex]y_1^{\prime\prime} = r - py_1^{\prime} - qy_1[/itex] in we get:

[itex] - r \int \frac{ry_2}{W}\;dx + \int \frac{ry_1}{W}\;dx \left( y_2^{\prime\prime} + py_2^{\prime} + qy_2 \right)[/itex]

Doing the same for [itex]y_2^{\prime\prime} = r - py_2^{\prime} - qy_2[/itex] we get:

[itex] - r \int \frac{ry_2}{W}\;dx + r \int \frac{ry_1}{W}\;dx [/itex]
 
  • #12
EDIT:

[itex]y_1[/itex] and [itex]y_2[/itex] are linearly independent solutions of the homogeneous equation [itex]y^{\prime\prime} + py^{\prime} + qy = 0[/itex]. (not the inhomogeneous one as in my last post).

So this would make [itex]-\int \frac{ry_2}{W}\;dx \left( y_1^{\prime\prime} + py_1^{\prime} + qy_1 \right) + \int \frac{ry_1}{W}\;dx \left( y_2^{\prime\prime} + py_2^{\prime} + qy_2 \right)[/itex]

equal 0 which isn't r ...
 
  • #13
Ted123 said:

Homework Statement



Show (by substituting it directly into the differential equation) that

[itex]\displaystyle y_p = y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx[/itex]

is a particular solution of [itex]y'' + p(x)y' + q(x)y = r(x)[/itex].

Homework Equations



[itex]W[/itex] is the Wronskian [itex]y_1 y_2^{\prime} - y_2 y_1^{\prime}[/itex]

The Attempt at a Solution



How do I sub [itex]y_p[/itex] into the DE?

Bad notation is getting in your way. You need a different integration variable, so instead of having dx inside the integral you should have, for example, dt; that is, write
[itex]\displaystyle y_p(x) = y_2(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt - y_1(x) \int_{a}^{x} \frac{r(t)y_2(t)}{W(t)}\,dt [/itex]
Now you can see that the derivative of the first term is
[itex] \displaystyle y_2'(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt + y_2(x) \frac{r(x)y_1(x)}{W(x)} , [/itex]
etc.

RGV
 
Last edited:
  • #14
Ray Vickson said:
Bad notation is getting in your way. You need a different integration variable, so instead of having dx inside the integral you should have, for example, dt; that is, write
[itex]\displaystyle y_p(x) = y_2(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt - y_1(x) \int_{a}^{x} \frac{r(t)y_2(t)}{W(t)}\,dt [/itex]
Now you can see that the derivative of the first term is
[itex] \displaystyle y_2'(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt + y_2(x) \frac{r(x)y_1(x)}{W(x)} , [/itex]
etc.

RGV

Even with different variables, it still leads to this:

[itex]\left( y_2^{\prime\prime} + py_2^{\prime} + qy_2 \right)\int \frac{ry_1}{W}\;dx - \left( y_1^{\prime\prime} + py_1^{\prime} + qy_1 \right)\int \frac{ry_2}{W}\;dx [/itex]

which I need to show equals [itex]r[/itex].

There must be something with those DEs multiplying each integral that means something
 
  • #15
I get the above + r. The coefficients multiplying your two integrals are zero, since y1 and y2 satisfy the homogeneous DE.

RGV
 
  • #16
Ray Vickson said:
I get the above + r. The coefficients multiplying your two integrals are zero, since y1 and y2 satisfy the homogeneous DE.

RGV

Can I ask how you managed to get that +r ?

[itex]\displaystyle y_p^{\prime} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx[/itex]

[itex]\displaystyle y_p^{\prime\prime} = y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx[/itex]

So subbing these into [itex]y_p^{\prime\prime} + py_p^{\prime} + qy_p[/itex] gives:

[itex]\displaystyle y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx + p(x) \left( y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx \right) + q(x) \left( y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx \right)[/itex]

[itex]= \left( y_2^{\prime\prime} + py_2^{\prime} + qy_2 \right)\int \frac{ry_1}{W}\;dx - \left( y_1^{\prime\prime} + py_1^{\prime} + qy_1 \right)\int \frac{ry_2}{W}\;dx [/itex]
 
  • #17
Ted123 said:
Can I ask how you managed to get that +r ?

[itex]\displaystyle y_p^{\prime} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx[/itex]

[itex]\displaystyle y_p^{\prime\prime} = y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx[/itex]

So subbing these into [itex]y_p^{\prime\prime} + py_p^{\prime} + qy_p[/itex] gives:

[itex]\displaystyle y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx + p(x) \left( y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx \right) + q(x) \left( y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx \right)[/itex]

[itex]= \left( y_2^{\prime\prime} + py_2^{\prime} + qy_2 \right)\int \frac{ry_1}{W}\;dx - \left( y_1^{\prime\prime} + py_1^{\prime} + qy_1 \right)\int \frac{ry_2}{W}\;dx [/itex]

You have the wrong [tex]y_p''[/tex]. Remember: [tex] \displaystyle \frac{d}{dx} f(x) \int_{a}^x g(y) \, dy = f'(x) \int_{a}^x g(y) \, dy + f(x) g(x) [/tex]

RGV
 
  • #18
Ray Vickson said:
You have the wrong [tex]y_p''[/tex]. Remember: [tex] \displaystyle \frac{d}{dx} f(x) \int_{a}^x g(y) \, dy = f'(x) \int_{a}^x g(y) \, dy + f(x) g(x) [/tex]

RGV

Of course. The [itex]y_p^{\prime\prime}[/itex] has the +r as the W cancels! (Thanks a lot!)
 

1. What is a particular solution (yp) in differential equations?

A particular solution (yp) in differential equations is a specific solution that satisfies the given differential equation, along with any initial or boundary conditions. It is typically found by adding a complementary solution (yc) to the general solution of the differential equation.

2. How is a particular solution found in differential equations?

A particular solution is typically found by using the method of undetermined coefficients or the method of variation of parameters. These methods involve making an educated guess for the form of the particular solution and then solving for the unknown coefficients.

3. What role does the inhomogeneous term (r(x)) play in finding a particular solution?

The inhomogeneous term (r(x)) represents any external forces or inputs in the differential equation. It is used in finding the particular solution because it helps determine the form of the solution and the unknown coefficients needed to satisfy the equation.

4. Can a particular solution be unique in differential equations?

No, a particular solution is not necessarily unique in differential equations. This is because there can be multiple solutions that satisfy the given equation and initial or boundary conditions. However, the combination of a particular and complementary solution will always be unique.

5. Why is it important to find a particular solution in differential equations?

Finding a particular solution is important because it allows us to solve for a specific solution that satisfies the given equation and any initial or boundary conditions. This is useful in many applications, such as modeling physical systems or predicting future behavior of a system.

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