J's answer to P's Junior College 2 H2 Maths Singapore question.
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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
(Shown)
= 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
(Shown)
Date Posted:
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.
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.
Thank you very much for your help.
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