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Year 12s: what will be the first way you research your future options? >>

FP3 differential equations

Hey, can someone help me with this FP3 question?

Show that the substitution

y = usinx

transforms the differential equation

sin^{2}x\frac{d^{2}y}{dx^{2}} -2sinxcosx\frac{dy}{dx} + 2y = 2sin^{4}xcosx

into

\frac{d^{2}u}{dx^{2}} + u = sin2x

Thanks!

(edited 10 years ago)

Original post by Cubic

Hey, can someone help me with this FP3 question?

Show that the substitution

y = usinx

transforms the differential equation

sin^{2}x\frac{d^{2}y}{dx^{2}} -2sinxcosx\frac{dy}{dx} + 2y = 2sin^{4}xcosx

into

\frac{d^{2}u}{dx^{2}} + u = sin2x

Thanks!

~~Haven't tried it yet, but~~ note that u is a function of x and sin(x) is a function of x, so try differentiating y with respect to x using the ~~chain~~ product rule, and doing it again for the second derivative. Then sub it all in, and try and simplify the DE.

Edit: Just tried it, works out fine.

(edited 10 years ago)

Original post by Cubic

Hey, can someone help me with this FP3 question?

Show that the substitution

y = usinx

transforms the differential equation

sin^{2}x\frac{d^{2}y}{dx^{2}} -2sinxcosx\frac{dy}{dx} + 2y = 2sin^{4}xcosx

into

\frac{d^{2}u}{dx^{2}} + u = sin2x

Thanks!

What have you tried so far?

I would expect that if you just "follow your nose" and keep applying the product rule to differentiating y = usinx the answer would hopefully drop out.

OP

Original post by Blazy

Haven't tried it yet, but note that u is a function of x and sin(x) is a function of x, so try differentiating y with respect to x using the chain rule, and doing it again for the second derivative.

Like this?

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}

Not really sure what I'm doing. :s-smilie:

:s-smilie:

Original post by Cubic

Like this?

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}

Not really sure what I'm doing. :s-smilie:

:s-smilie:

Ah crap, I meant product rule not chain rule! (Corrected in first post) So sorry!

OP

u = u, v = sinx
u' = 1, v' = cosx

\frac{dy}{dx} = sinx + ucosx

u = u, v = cos x
u' = 1, v' = -sinx

\frac{d^{2}y}{dx^{2}} = cos x + cos x - usinx = 2cosx -usinx

Is that ok so far?

Original post by Cubic

u = u, v = sinx
u' = 1, v' = cosx

\frac{dy}{dx} = sinx + ucosx

u = u, v = cos x
u' = 1, v' = -sinx

\frac{d^{2}y}{dx^{2}} = cos x + cos x - usinx = 2cosx -usinx

Is that ok so far?

Not quite. It seems you're a bit confused about the

u

bit. Think of it as a function of x, i.e.

y = u(x)\sin(x)

. Since u is a function of x, when we differentiate it we get

\frac{du}{dx}

(Don't think I can reveal much more, otherwise I'll be doing it for you!)

OP

\frac{dy}{dx} = \frac{du}{dx}sinx + ucosx

\frac{d^{2}y}{dx^{2}} = \frac{d^{2}u}{dx^{2}}sinx + 2 \frac{du}{dx}cosx - usinx

Like this?

Original post by Cubic

\frac{dy}{dx} = \frac{du}{dx}sinx + ucosx

\frac{d^{2}y}{dx^{2}} = \frac{d^{2}u}{dx^{2}}sinx + 2 \frac{du}{dx}cosx - usinx

Like this?

Yup. Sub it all back in and see how it goes :smile:

:smile:

OP

sin^{2}x(\frac{d^{2}u}{dx^{2}}sinx + 2\frac{du}{dx}cosx - usinx) -2sinxcosx(\frac{du}{dx}sinx + ucosx) +2usinx = 2sin^{4}xcosx

\frac{d^{2}u}{dx^{2}}sin^{3}x + 2\frac{du}{dx}cosxsin^{2}x - usin^{3}x - 2\frac{du}{dx}cosxsin^{2}x -2usinxcos^{2}x +2usinx = 2sin^{4}xcosx

\frac{d^{2}u}{dx^{2}}sin^{3}x - usin^{3}x - 2usinxcos^{2}x +2usinx = 2sin^{4}xcosx

\frac{d^{2}u}{dx^{2}}sin^{2}x - usin^{2}x -2ucos^{2}x + 2u = sin^{2}x(sin2x)

:confused:

Original post by Cubic

sin^{2}x(\frac{d^{2}u}{dx^{2}}sinx + 2\frac{du}{dx}cosx - usinx) -2sinxcosx(\frac{du}{dx}sinx + ucosx) +2usinx = 2sin^{4}xcosx

\frac{d^{2}u}{dx^{2}}sin^{3}x + 2\frac{du}{dx}cosxsin^{2}x - usin^{3}x - 2\frac{du}{dx}cosxsin^{2}x -2usinxcos^{2}x +2usinx = 2sin^{4}xcosx

\frac{d^{2}u}{dx^{2}}sin^{3}x - usin^{3}x - 2usinxcos^{2}x +2usinx = 2sin^{4}xcosx

\frac{d^{2}u}{dx^{2}}sin^{2}x - usin^{2}x -2ucos^{2}x + 2u = sin^{2}x(sin2x)

:confused:

Substitute:

cos^2x = 1-sin^2x

OP

Original post by SherlockHolmes

Substitute:

cos^2x = 1-sin^2x

That was so beautiful.

Thank you all so much!

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