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# P4 - Differential Equations watch

1. Could someonexplain to me how they got from the "*****" to the next stage of this proof? Thanks

Q.Given that x=e^u, where u is a function of x, show that:
x^2.d^2.y/dx^2 = [d^2.y/du^2] - dy/du

A.
d^2.y/du^2 = [d/du].(dy/du)
= [d/du].(e^u.dy/dx) *****
= e^u.dy/dx + e^u.[d^2.y/dx^2]. [dx/du]
= dy/du + x^2.[d^2.y/dx^2]

Since dx/du = e^u and e^u=x

x^2.[d^2.y/dx^2] = [d^2.y/du^2] - dy/du

(Sorry the layouts not v.easy to understand - I think UKL should have special function buttons for us Mathematicians - or at least for those of us trying to be mathematicians!!!!) Thanks
2. A.
d^2.y/du^2 = [d/du].(dy/du)
= [d/du].(e^u.dy/dx) *****

ok, from this point, you're simply using the product rule. viz.

d(uv)/dx = u.dv/dx + v.du/dx

( the u above is different from the u that ur using

so,
[d/du].(e^u.dy/dx) = (e^u).[d/du](dy/dx) + (dy/dx).[d/du](e^u)
[d/du].(e^u.dy/dx) = (e^u).d²y/dx².dx/du + dy/dx.e^u
[d/du].(e^u.dy/dx) = e^u.dy/dx + (e^u).d²y/dx².dx/du

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