Transformation

**Erghhh** · April 3rd 2009, 06:20 AM

(2x)d2t/dx2 - 6(dx/dt)^2 = x^2 - 3x^4

show that the substitution y=1/x^2 transforms this differential eqn into d2y/dt2 + y = 3

So i've work out

-2/x^3(d2x/dt2) + 6/x^4(dx/dt)

i think the latter part of this eqn is wrong as the mark scheme says it's 6/x^4(dx/dt)^2

Can anyone explain this?

**Danny** · April 3rd 2009, 06:54 AM

Originally Posted by Erghhh

(2x)d2t/dx2 - 6(dx/dt)^2 = x^2 - 3x^4

show that the substitution y=1/x^2 transforms this differential eqn into d2y/dt2 + y = 3

So i've work out

-2/x^3(d2x/dt2) + 6/x^4(dx/dt)

i think the latter part of this eqn is wrong as the mark scheme says it's 6/x^4(dx/dt)^2

Can anyone explain this?

Sure

$y = x^{-2}$ so $\frac{dy}{dt} = -2x^{-3} \frac{dx}{dt}$ . Thus,
$\frac{d^2y}{dt^2} = \frac{d}{dt} \left(-2x^{-3} \frac{dx}{dt}\right) = 6x^{-4} \underbrace{\frac{dx}{dt} \cdot \frac{dx}{dt}}_{ \dot{x}^2 } -2x^{-3} \frac{d^2x}{dt^2}$

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