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junior college 2 | H2 Maths
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junior college 2 chevron_right H2 Maths chevron_right Singapore

Need help for q3

Date Posted: 3 years ago
Views: 240
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Using product rule, d/dt (t dx/dt)
= d/dt (t) dx/dt + t d/dt (dx/dt)
= (1) dx/dt + t d²x/dt²
= t d²x/dt² + dx/dt
So integrating this will give you back t dx/dt
Next,
t d²x/dt² + dx/dt = 4t - 1
Integrate both sides with respect to t,
∫ (t d²x/dt² + dx/dt) dt = ∫ (4t - 1) dt
t dx/dt = 2t² - t + C, C is a constant
dx/dt = 2t - 1 + C/t
Integrate both sides with respect to t again,
∫ (dx/dt) dt = ∫ (2t - 1 + C/t) dt
x = t² - t + C lnt + D
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J
J's answer
1024 answers (A Helpful Person)
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J
J
3 years ago
In case you're unsure of the last few steps,

For a function x = f(t), differentiating x with respect to t gives :

d/dt (x) = f'(t)

dx/dt = f'(t)

So integrating ,

∫ (dx/dt) dt = ∫ f'(t) dt

It will just give you back x = f(t) , but with a constant since we were not sure if there was any constant to start with if we are only given dx/dt = f'(t) at the beginning.


Eg. x = cos t

d/dt (x) = d/dt (cos t)

dx/dt = -sin t

So,

∫ (dx/dt) dt = ∫ -sin t dt
x = cos t + C, C is a arbitrary constant.
P
P
3 years ago
Thank you very much for your help.