Help with Edexcel FP2 Second Order Differential Equation substitution question

Use the substitution z = sinx to transform the differential equation

cosx(d^2y/dx^2) + sinx(dy/dx) - 2ycos^3x = 2cosx^5x

into the equation (d^2y/dz^2) - 2y = 2(1 - z^2)

This isn't really about how to do such a question in general, more a probably embarrassing gap in my understanding of differentiation.

So I realise that all terms in x must be transformed. Focusing first on the non-differential ones

sinx is easy enough, sinx = z.
cosx is (1-z^2)^(1/2)
It follows that cos^3x and cos^5x are (1 - z^2)^(3/2) and (1 - z^2)^(5/2) respectively.

dy/dx will = dy/dz * dz/dx. But we have z as a function of x, z = sinx and so dz/dx = cosx = (1 - z^2)^(1/2)

Therefore dy/dx = dy/dz(1 - z^2)^(1/2).

But now the problem comes with d^2y/dx^2.
How do I differentiate dy/dz with respect to x? dy/(dzdx) doesn't really seem to get me anywhere if that's the technical representation..
So I tried d^2y/dx^2 = (d^2y/dz^2)(dz^2/dx^2), but knowing that the relationship dy/dx = dy/du * du/dx is a consequence of the chain rule, rather than anything to do with fractions or the like, I doubt/don't know whether this extends to the second order differentials. After working through the consequences of said assumption (namely, d^2y/dx^2 = -sinx(d^2y/dz^2)), the needed equation did not drop out.

So yeah, assuming that that is incorrect - if not, then I guess I must have gone wrong afterwards, and if so that can probably be corrected by more thorough working, so please tell me if that's the case - how else could I get d^2y/dx^2 into terms of just y and z? And even if it isn't the best way to do it, how exactly would you differentiate (dy/dz) with respect to x?

(edited 9 years ago)

Reply 1

ztibor

10

Original post by 1 8 13 20 42

Use the substitution z = sinx to transform the differential equation

cosx(d^2y/dx^2) + sinx(dy/dx) - 2ycos^3x = 2cosx^5x

into the equation (d^2y/dx^2) - 2y = 2(1 - z^2)

This isn't really about how to do such a question in general, more a probably embarrassing gap in my understanding of differentiation.

So I realise that all terms in x must be transformed. Focusing first on the non-differential ones

sinx is easy enough, sinx = z.
cosx is (1-z^2)^(1/2)
It follows that cos^3x and cos^5x are (1 - z^2)^(3/2) and (1 - z^2)^(5/2) respectively.

dy/dx will = dy/dz * dz/dx. But we have z as a function of x, z = sinx and so dz/dx = cosx = (1 - z^2)^(1/2)

Therefore dy/dx = dy/dz(1 - z^2)^(1/2).

But now the problem comes with d^2y/dx^2.
How do I differentiate dy/dz with respect to x? dy/(dzdx) doesn't really seem to get me anywhere if that's the technical representation..

USe the chain rule

\displaystyle \frac{d^2y}{dx^2}=\frac{dy'}{dx}=\frac{d}{dx}\left (\frac{dy}{dz}\cdot \sqrt{1-z^2}\right ) =

\displaystyle =\frac{d^2y}{dz^2}\cdot \frac{dz}{dx}\cdot \sqrt{1-z^2} -\frac{z}{ \sqrt{1-z^2}}\cdot \frac{dz}{dx}

Reply 2

davros

16

Original post by 1 8 13 20 42

Use the substitution z = sinx to transform the differential equation

cosx(d^2y/dx^2) + sinx(dy/dx) - 2ycos^3x = 2cosx^5x

into the equation (d^2y/dz^2) - 2y = 2(1 - z^2)

This isn't really about how to do such a question in general, more a probably embarrassing gap in my understanding of differentiation.

So I realise that all terms in x must be transformed. Focusing first on the non-differential ones

sinx is easy enough, sinx = z.
cosx is (1-z^2)^(1/2)
It follows that cos^3x and cos^5x are (1 - z^2)^(3/2) and (1 - z^2)^(5/2) respectively.

dy/dx will = dy/dz * dz/dx. But we have z as a function of x, z = sinx and so dz/dx = cosx = (1 - z^2)^(1/2)

Therefore dy/dx = dy/dz(1 - z^2)^(1/2).

But now the problem comes with d^2y/dx^2.
How do I differentiate dy/dz with respect to x? dy/(dzdx) doesn't really seem to get me anywhere if that's the technical representation..
So I tried d^2y/dx^2 = (d^2y/dz^2)(dz^2/dx^2), but knowing that the relationship dy/dx = dy/du * du/dx is a consequence of the chain rule, rather than anything to do with fractions or the like, I doubt/don't know whether this extends to the second order differentials. After working through the consequences of said assumption (namely, d^2y/dx^2 = -sinx(d^2y/dz^2)), the needed equation did not drop out.

So yeah, assuming that that is incorrect - if not, then I guess I must have gone wrong afterwards, and if so that can probably be corrected by more thorough working, so please tell me if that's the case - how else could I get d^2y/dx^2 into terms of just y and z? And even if it isn't the best way to do it, how exactly would you differentiate (dy/dz) with respect to x?

The bold bit certainly isn't true.

You basically want to be using dw/dx = dw/dz * dz/dx which is true for any function w - in particular w can be dy/dx which you should already have worked out in terms of z.

Reply 3

KeithHayward

2

Original post by 1 8 13 20 42

Use the substitution z = sinx to transform the differential equation

cosx(d^2y/dx^2) + sinx(dy/dx) - 2ycos^3x = 2cosx^5x

into the equation (d^2y/dz^2) - 2y = 2(1 - z^2)

This isn't really about how to do such a question in general, more a probably embarrassing gap in my understanding of differentiation.

So I realise that all terms in x must be transformed. Focusing first on the non-differential ones

sinx is easy enough, sinx = z.
cosx is (1-z^2)^(1/2)
It follows that cos^3x and cos^5x are (1 - z^2)^(3/2) and (1 - z^2)^(5/2) respectively.

dy/dx will = dy/dz * dz/dx. But we have z as a function of x, z = sinx and so dz/dx = cosx = (1 - z^2)^(1/2)

Therefore dy/dx = dy/dz(1 - z^2)^(1/2).

But now the problem comes with d^2y/dx^2.
How do I differentiate dy/dz with respect to x? dy/(dzdx) doesn't really seem to get me anywhere if that's the technical representation..
So I tried d^2y/dx^2 = (d^2y/dz^2)(dz^2/dx^2), but knowing that the relationship dy/dx = dy/du * du/dx is a consequence of the chain rule, rather than anything to do with fractions or the like, I doubt/don't know whether this extends to the second order differentials. After working through the consequences of said assumption (namely, d^2y/dx^2 = -sinx(d^2y/dz^2)), the needed equation did not drop out.

So yeah, assuming that that is incorrect - if not, then I guess I must have gone wrong afterwards, and if so that can probably be corrected by more thorough working, so please tell me if that's the case - how else could I get d^2y/dx^2 into terms of just y and z? And even if it isn't the best way to do it, how exactly would you differentiate (dy/dz) with respect to x?