Question

4. a) A wet porus substance in the open air loses its moisture at a rate proportional to the moisture
content. If a sheet hung in the wind loses half its moisture during the first hour, then find the
time when it has lost 95% moisture provided the weather conditions remain the same.

b) If the interest is compounded continuously, the amount of money invested increases at a rate
proportional to its size. Let Rs. 10,000 be invested at 10% compounded continuously. Then in
how many years will the original investment double itself ?

c) Solve: dy/dx + (x/1-x^2) y = xy, y(0)=1.

Answer 1

Answer on Question #50897-Math-Differential Calculus-Equations

a) A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half its moisture during the first hour, then find the time when it has lost 95% moisture provided the weather conditions remain the same.

b) If the interest is compounded continuously, the amount of money invested increases at a rate proportional to its size. Let Rs. 10,000 be invested at 10% compounded continuously. Then in how many years will the original investment double itself?

c) Solve: $dy/dx + (x/1 - x^2) y = xy$, $y(0) = 1$.

Solution

a)

$$\frac{dM}{dt} = -kM \rightarrow M = M_0 e^{-kt}.$$ $$M(1 \text{ hour}) = \frac{1}{2} M_0 \rightarrow \frac{1}{2} = e^{-k} \rightarrow k = \ln 2 \text{ hours}^{-1}.$$ $$M(T) = M_0 (1 - 0.95) = 0.05 M_0 = M_0 e^{-kT}.$$ $$T = \frac{1}{k} \ln 20 = \frac{\ln 20}{\ln 2} = 4.3 \text{ hours}.$$

Answer: 4.3 hours.

b) $r = 0.1$.

$$R = R_0 e^{rt}.$$ $$R(T) = 2 R_0 = R_0 e^{rT}.$$ $$T = \frac{1}{r} \ln 2 = \frac{\ln 2}{0.1} = 6.9 \text{ years}.$$

Answer: 6.9 years.

c)

$$\frac{dy}{dx} + \left(\frac{x}{1 - x^2}\right) y = x \sqrt{y}, \quad y(0) = 1.$$ $$t^2 = y \rightarrow 2t \frac{dt}{dx} + \left(\frac{x}{1 - x^2}\right) t^2 = xt.$$ $$2 \frac{dt}{dx} + \left(\frac{x}{1 - x^2}\right) t = x.$$

General solution:

$$2 \frac{dt}{dx} + \left(\frac{x}{1 - x^2}\right) t = 0 \rightarrow \frac{2dt}{t} = -\left(\frac{x dx}{1 - x^2}\right) = \frac{1}{2} \frac{d(1 - x^2)}{1 - x^2} \rightarrow \ln \frac{t^2}{t_0^2} = \frac{1}{2} \ln (1 - x^2).$$ $$t ^ {2} = t _ {o} ^ {2} \sqrt {1 - x ^ {2}} \rightarrow t = t _ {0} \sqrt [ 4 ]{1 - x ^ {2}}.$$

Partial solution is

$$t = - \frac {1}{3} (1 - x ^ {2}).$$

Thus

$$y = \left(t _ {0} \sqrt [ 4 ]{1 - x ^ {2}} - \frac {1}{3} (1 - x ^ {2})\right) ^ {2}, y (0) = 1.$$ $$1 = \left(t _ {0} - \frac {1}{3}\right) ^ {2} \rightarrow t _ {0} = \frac {4}{3}.$$

Answer: $\left(\frac{4}{3}\sqrt[4]{1 - x^2} -\frac{1}{3}\big(1 - x^2\big)\right)^2.$

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