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!)