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The solution of the differential equation $\dfrac{dy}{dx} = (x-y)^2$, subject to $y(1)=1$, is:
Let $v = x - y$, so $\frac{dv}{dx} = 1 - \frac{dy}{dx} = 1 - v^2$.
Separate: $\frac{dv}{1-v^2} = dx \Rightarrow \frac{1}{2}\ln\left|\frac{1+v}{1-v}\right| = x + C$
i.e. $\ln\left|\frac{1+x-y}{1-x+y}\right| = 2x + C'$
At $(1,1)$: $v=0$, $\ln(1)=2+C' \Rightarrow C'=-2$
Final: $-\ln\left|\dfrac{1-x+y}{1+x-y}\right| = 2(x-1)$ which matches option B.
Which one of the following qualifies as a capital expenditure?
Capital expenditure is spending that improves, extends the life of, or adds value to a long-term asset. Renovating and upgrading a building increases its value and useful life — this is capital expenditure. The other options are all revenue expenditures: carriage outwards, legal recovery fees, and wage bonuses are period costs that appear on the income statement and do not create lasting assets.
- Spring constant $k$: $F = kx \Rightarrow [k] = \frac{[F]}{[x]} = \frac{MLT^{-2}}{L} = MT^{-2} = M^1L^0T^{-2}$ → (3)
- Pascal (pressure): $[P] = \frac{Force}{Area} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} = M^1L^{-1}T^{-2}$ → (4)
- Hertz (frequency): $[f] = T^{-1} = M^0L^0T^{-1}$ → (2)
- Joule (energy): $[E] = ML^2T^{-2} = M^1L^2T^{-2}$ → (1)
Correct match: A-3, B-4, C-2, D-1
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