Particular Sub y_p into DE for r(x)

[Ted123]

Homework Statement

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

$\displaystyle y_p = y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx$

is a particular solution of $y'' + p(x)y' + q(x)y = r(x)$.

Homework Equations

$W$ is the Wronskian $y_1 y_2^{\prime} - y_2 y_1^{\prime}$

The Attempt at a Solution

How do I sub $y_p$ into the DE?

 

[micromass]

Hi Ted

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

 

[Ted123]

Hi Ted

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

I can't see how to differentiate $y_p$ and certainly $y_p^{\prime}$ though.

For instance, how do I find:

$\displaystyle \frac{d}{dx} \left( y_2 \int \frac{ry_1}{W}\;dx \right)$ ?

I know

$\frac{d}{dx} \left( \int \frac{ry_1}{W}\;dx \right) = \frac{ry_1}{W}$

but $y_2$ is a function of x, and it is multiplied by that integral...

 

[micromass]

Ah yes. There, you have to differentiate a product of two functions. So you need to use the product rule:

$$(fg)^\prime=f^\prime g+fg^\prime$$

 

[Ted123]

Ah yes. There, you have to differentiate a product of two functions. So you need to use the product rule:

$$(fg)^\prime=f^\prime g+fg^\prime$$

I get:

$\displaystyle y_p^{\prime} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx$

$\displaystyle y_p^{\prime\prime} = y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx$

So:

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

should equal r but does it?

 

[micromass]

You didn't exactly follow the product rule did you??

You should have done

$$\left(y_2\int\frac{ry_1}{W}\right)^\prime=y_2^\prime\int{\frac{ry_1}{W}}+y_2\frac{ry_1}{W}$$

Now, if you differentiate like that, then what is $$y_p$$ then?

 

[Ted123]

You didn't exactly follow the product rule did you??

You should have done

$$\left(y_2\int\frac{ry_1}{W}\right)^\prime=y_2^\prime\int{\frac{ry_1}{W}}+y_2\frac{ry_1}{W}$$

Now, if you differentiate like that, then what is $$y_p$$ then?

I did follow the product rule:

$\displaystyle y_p = y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx$

so differentiating the first term gives what you have.

Then differentiating the 2nd term $- y_1 \int \frac{ry_2}{W}\;dx$ gives:

$- y_1^{\prime} \int \frac{ry_2}{W}\;dx - \frac{ry_1y_2}{W}$

so adding the 2 together gives:

$y_p^{\prime} = y_2^\prime\int{\frac{ry_1}{W}\;dx}+\frac{ry_1y_2}{W} - y_1^{\prime} \int \frac{ry_2}{W}\;dx - \frac{ry_1y_2}{W} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx$

 

[micromass]

Ah, yes, you are correct!
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?)

 

[Ted123]

Ah, yes, you are correct!
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?)

$y_1(x)$ and $y_2(x)$ are independent solutions of

$y^{\prime\prime} + p(x) y^{\prime} + q(x) y =0$.

 

[grey_earl]

So you can substitute for $$y_1''(x)$$ in your equation above.

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

 

[Ted123]

So you can substitute for $$y_1''(x)$$ in your equation above.

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

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

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

to equal $r(x)$.

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

If we substitute $y_1^{\prime\prime} = r - py_1^{\prime} - qy_1$ in we get:

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

Doing the same for $y_2^{\prime\prime} = r - py_2^{\prime} - qy_2$ we get:

$- r \int \frac{ry_2}{W}\;dx + r \int \frac{ry_1}{W}\;dx$

 

[Ted123]

EDIT:

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

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

equal 0 which isn't r ...

 

[Ray Vickson]

Homework Statement

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

$\displaystyle y_p = y_2 \int \frac{ry_1}{W}\;dx - y_1 \int \frac{ry_2}{W}\;dx$

is a particular solution of $y'' + p(x)y' + q(x)y = r(x)$.

Homework Equations

$W$ is the Wronskian $y_1 y_2^{\prime} - y_2 y_1^{\prime}$

The Attempt at a Solution

How do I sub $y_p$ 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
$\displaystyle y_p(x) = y_2(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt - y_1(x) \int_{a}^{x} \frac{r(t)y_2(t)}{W(t)}\,dt$
Now you can see that the derivative of the first term is
$\displaystyle y_2'(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt + y_2(x) \frac{r(x)y_1(x)}{W(x)} ,$
etc.

RGV

 

[Ted123]

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
$\displaystyle y_p(x) = y_2(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt - y_1(x) \int_{a}^{x} \frac{r(t)y_2(t)}{W(t)}\,dt$
Now you can see that the derivative of the first term is
$\displaystyle y_2'(x) \int_{a}^{x} \frac{r(t)y_1(t)}{W(t)}\,dt + y_2(x) \frac{r(x)y_1(x)}{W(x)} ,$
etc.

RGV

Even with different variables, it still leads to this:

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

which I need to show equals $r$.

There must be something with those DEs multiplying each integral that means something

 

[Ray Vickson]

I get the above + r. The coefficients multiplying your two integrals are zero, since y1 and y2 satisfy the homogeneous DE.

RGV

 

[Ted123]

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 ?

$\displaystyle y_p^{\prime} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx$

$\displaystyle y_p^{\prime\prime} = y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx$

So subbing these into $y_p^{\prime\prime} + py_p^{\prime} + qy_p$ gives:

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

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

 

[Ray Vickson]

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

$\displaystyle y_p^{\prime} = y_2^{\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime} \int \frac{ry_2}{W}\;dx$

$\displaystyle y_p^{\prime\prime} = y_2^{\prime\prime} \int \frac{ry_1}{W}\;dx - y_1^{\prime\prime} \int \frac{ry_2}{W}\;dx$

So subbing these into $y_p^{\prime\prime} + py_p^{\prime} + qy_p$ gives:

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

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

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

RGV

 

[Ted123]

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

RGV

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