# Edexcel P5 Style Paper HeinnemanWatch

This discussion is closed.
#1
From my experience, these papers tend to be more difficult than past papers, but I'm trying this one anyway. I'm having a bit of trouble with the first(!) question:

" Given y = xsinh x, show that y(d²y/dx²) - (dy/dx)² = y²/x "

So, I started by taking logs and then differentiating both sides with respect to x twice, then multiplying through by y². This left me with y(d²y/dx²) - (dy/dx)² = y²[(2/x)cosh x - (1/x²)sinhx + cosh x·ln x]

Now the left hand side's correct and I've got the y² on the right, but something tells me the rest of that stuff doesn't equal 1/x... I've obviously gone horribly wrong somewhere - anybody fancy having a go and telling me the right way to do it?
0
13 years ago
#2
(Original post by drw25)
From my experience, these papers tend to be more difficult than past papers, but I'm trying this one anyway. I'm having a bit of trouble with the first(!) question:

" Given y = xsinh x, show that y(d²y/dx²) - (dy/dx)² = y²/x "

So, I started by taking logs and then differentiating both sides with respect to x twice, then multiplying through by y². This left me with y(d²y/dx²) - (dy/dx)² = y²[(2/x)cosh x - (1/x²)sinhx + cosh x·ln x]

Now the left hand side's correct and I've got the y² on the right, but something tells me the rest of that stuff doesn't equal 1/x... I've obviously gone horribly wrong somewhere - anybody fancy having a go and telling me the right way to do it?
Mine says y=x[SUP]x and worked fine when I did the question. Perhaps your book suffers from a printing error?
0
#3
Hmm...looks like you're right - thanks! Mine was printed in 2001, they've probably fixed it since. I can do it no probs with x^x...
0
13 years ago
#4
Yeah; our teacher told us that was a printing error (mine read e^sinhx too...)
0
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