Answer to Question #107633 in Differential Equations for Nikesh gautam pandit ji

Question

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.

Answer 1

$dy/dx + (x/1-x^2)y = x√y$ ---(1)

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

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

$(1/\sqrt y)dy/dx + (x/(1-x^2))\sqrt y = x$

which is a standard Bernoulli's Differential equation which solved in the following standard manner:

Substitute $\sqrt y=t \implies 1/(2\sqrt y)*dy/dx=dt/dx$ we get;

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

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

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

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

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

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

Thus the solution is :

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

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

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

Putting t and u back; we get;

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

Using (2) we get;

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

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

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

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

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

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

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