Question #107633
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/1x2)y=xydy/dx + (x/1-x^2)y = x√y ---(1)

y(0)=1y(0)=1 ---(2)

Dividing (1) by y\sqrt y we get;

(1/y)dy/dx+(x/(1x2))y=x(1/\sqrt y)dy/dx + (x/(1-x^2))\sqrt 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\sqrt y=t \implies 1/(2\sqrt y)*dy/dx=dt/dx we get;

dt/dx+tx/(2(1x2))=x/2dt/dx+ tx/(2(1-x^2)) = x/2

which is of the form : dt/dx+tP(x)=Q(x)dt/dx+tP(x)=Q(x)

whose solution is given as : t(IF)=Q(IF)dxt(IF)=\int Q(IF)dx where IF=ePdxIF=e^{\int Pdx}

Here, P(x)=x/(2(1x2));Q(x)=x/2P(x)=x/(2(1-x^2)); Q(x) = x/2

Thus, IF=exdx/(2(1x2))=e2xdx/(4(1x2))IF=e^{\int xdx/(2(1-x^2))}=e^{\int 2xdx/(4(1-x^2))}

=e(ln(1x2))/4=(1x2)1/4=e^{-(ln(1-x^2))/4}=(1-x^2)^{-1/4}

Thus the solution is :

t(1x2)1/4=xdx/(2(1x2)1/4)=2xdx/(4(1x2)1/4)t(1-x^2)^{-1/4}=\int xdx/(2(1-x^2)^{1/4})=\int 2xdx/(4(1-x^2)^{1/4})

Put 1x2=u    2xdx=du1-x^2=u \implies -2xdx=du ;we get;

t(1x2)1/4=du/(4u1/4)=(1/3)u3/4+ct(1-x^2)^{-1/4}=-\int du/(4u^{1/4})=(-1/3)u^{3/4}+c

Putting t and u back; we get;

y(1x2)1/4=(1x2)3/4/3+c\sqrt y(1-x^2)^{-1/4}=-(1-x^2)^{3/4}/3+c

Using (2) we get;

1/(10)=1/3+c    c=4/31/(1-0)=-1/3+c \implies c=4/3

    y(1x2)1/4=(1x2)3/4/3+4/3\implies \sqrt y(1-x^2)^{-1/4}=-(1-x^2)^{3/4}/3+4/3


b) given : dm/dt=km-dm/dt=km where k is a constant of proportionality and m is moisture content at time t

Solving we get; m=moekt;m0m=m_oe^{-kt}; m_0 = moisture content at t=0

Also given : m(1hr.)=m0ek=m0/2    k=ln2m(1 hr.)=m_0e^{-k}=m_0/2\implies k=ln2

To find: t for which m(t)=0.05m0=m0ekt    kt=ln20m(t)=0.05 m_0=m_0e^{-kt} \implies kt=ln20

Thus, t=ln20/ln2=0.301hr.t=ln20/ln2=0.301 hr.


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