Solve the differential equationdy
dy/dx + (x/1-x^2)y = x√y, y(0)=1
b) 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.
1
Expert's answer
2020-04-09T14:17:59-0400
dy/dx+(x/1−x2)y=x√y ---(1)
y(0)=1 ---(2)
Dividing (1) by y we get;
(1/y)dy/dx+(x/(1−x2))y=x
which is a standard Bernoulli's Differential equation which solved in the following standard manner:
Substitute y=t⟹1/(2y)∗dy/dx=dt/dx we get;
dt/dx+tx/(2(1−x2))=x/2
which is of the form : dt/dx+tP(x)=Q(x)
whose solution is given as : t(IF)=∫Q(IF)dx where IF=e∫Pdx
Here, P(x)=x/(2(1−x2));Q(x)=x/2
Thus, IF=e∫xdx/(2(1−x2))=e∫2xdx/(4(1−x2))
=e−(ln(1−x2))/4=(1−x2)−1/4
Thus the solution is :
t(1−x2)−1/4=∫xdx/(2(1−x2)1/4)=∫2xdx/(4(1−x2)1/4)
Put 1−x2=u⟹−2xdx=du ;we get;
t(1−x2)−1/4=−∫du/(4u1/4)=(−1/3)u3/4+c
Putting t and u back; we get;
y(1−x2)−1/4=−(1−x2)3/4/3+c
Using (2) we get;
1/(1−0)=−1/3+c⟹c=4/3
⟹y(1−x2)−1/4=−(1−x2)3/4/3+4/3
b) given : −dm/dt=km where k is a constant of proportionality and m is moisture content at time t
Solving we get; m=moe−kt;m0 = moisture content at t=0
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