Question

1 H grams of artificial sugar in water are being converted into dextrose at a rate which is proportional to the square of the amount unconverted. Find the differential equation expressing the rate of conversion after v minutes given that s grams is converted in v minutes and c being the constant of proportionality.
(a)ds/dv = c(H−s)2
(b) ds/dv = c(H−s)^2
© ds/dv/= (s−H)^2
(d) dv/ds/= c(H−s)
2 A vehicle of mass m moves along a straight line ( the – axis) while subject to a force indirectly proportional to its displacement x from a fixed point O in its path and 2) a resisting force proportional to its acceleration. Express the total force as a differential equation.

(a) 7md^2/xdt^2=k1/t − k2dx/dt
(b) Md^2x/dt^2=−k1/t−k2d^2x/dt^2
© m d^2x/dt^2=−k1/x−k2d^2x/dt^2
(d) M d^−2x/dt^2=−2 k1/x − k2 d^2x/dt^2

Answer 1

Answer on Question #59053 – Math – Differential Equations

Question

1. H grams of artificial sugar in water are being converted into dextrose at a rate which is proportional to the square of the amount unconverted. Find the differential equation expressing the rate of conversion after v minutes given that s grams is converted in v minutes and c being the constant of proportionality.

(a) ds/dv = c(H-s)²

(b) ds/dv = c(H-s)^2

(c) ds/dv/= (s-H)^2

(d) dv/ds/= c(H-s)

Solution

Let amount converted be s, amount unconverted be H-s. Then $\frac{ds}{dv} = c(H - s)^2$.

Answer: (b) $\frac{ds}{dv} = c(H - s)^2$.

Question

2. A vehicle of mass m moves along a straight line (the - axis) while subject to a force indirectly proportional to its displacement x from a fixed point 0 in its path and 2) a resisting force proportional to its acceleration. Express the total force as a differential equation.

(a) $7md^2/xdt^2 = k1/t - k2dx/dt$

(b) $Md^2x/dt^2 = -k1/t - k2d^2x/dt^2$

(c) $m d^2x/dt^2 = -k1/x - k2d^2x/dt^2$

(d) $M d^2x/dt^2 = -2 k1/x - k2 d^2x/dt^2$

Solution

$$F = ma.$$ $$F = -\frac{k_1}{x} - k_2 a; \quad a = \frac{d^2x}{dt^2}.$$

Thus, $m \frac{d^2x}{dt^2} = -\frac{k_1}{x} - k_2 \frac{d^2x}{dt^2}$.

Answer: (c) $m \frac{d^2x}{dt^2} = -\frac{k_1}{x} - k_2 \frac{d^2x}{dt^2}$.

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