Examples : Particular solutions Video Lecture | Study Mathematics (Maths) Class 12 - JEE | Best Video for JEE

hello friends this video on differential
equation part 11 is brought you by exam
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watching this video please make sure
that you have watched part 1 2 part then
the volume of a spherical balloon being
inflated changes at a constant rate
initial radius is 3 unit and after 3
seconds it is season find the radius of
the balloon operty set so the question
says understand this question please
play attention it certification the
volume of a spherical balloon is
changing at a constant rate so the first
question we get is DV by DT is some
constant that is the first line correct
and we know is nothing but 4 by 3 PI R
for the spherical but correct also we
had told that initially that means T is
equal to 0 r is equal to 3 T is equal to
0 other than you in after 3 seconds T is
equal to 3 R is what this data we have
we have to find the ball radius act
please go to T what does the radius this
is what we have to find correct
understand does the question say is the
volume of this balloon is changing at a
constant rate that is DV by DT is
constant and V is equal to 4 by 3 PI R
QP know also we had told that initially
the dusties will be 0 r is equal to 3
and T is equal to 3 r is 1 to 6 so we
had told that DV by DT is equal to K
let's put this equation D V is nothing
but d of 4 by 3 PI R cube by DT is equal
to some constant correct so let's use
differential equation to solve this so
this becomes let's differentiate there
this Kelvin the by 3 PI R cube becomes 3
R square
right three R Square D apparently this
is pretty key correct that's what you
get three three cancels so what do we
get for pi is again constant integration
of R squared D R is equal to integration
of DT because we have got this this is
nothing but a differential equation
correct differential equation of order
one degree one so I got this
so if you solve this further this
becomes 4 PI R square becomes R cubed by
3 is equal to KT plus that means I got
an equation where I have this radius in
terms of T what does that find the value
of K and C because we have two variables
it's put the value of T is equal to 0 is
equal to 3 so this becomes full PI R a R
is equal to 0 0 sorry R is go to 3 3
cubed by 3 T is equal to 0 0 plus C
correct so you got C is equal to this
becomes 9 94 36 5 so I got the value of
C I have to mind well of K now let's put
this value of T is equal to 3 hours / 6
this becomes over PI R is equal to 6 Q
by 3 is equal to K need to find P is
equal to 3 plus C is at the six five
good so if you solve this you get K as
this becomes 6 into 6 30 36 into 6 0 8
into 4 288 to make 8 minus 36 plus 2 50
to 250 2 by 3 is 2 50 by 3 PI that is 84
5 correct you get K is equal to 84 5 k
is equal to 84 5 so I have got C a hot
key
my equation general equation turns out
the bell put this value here only space
here right so I get my 4 pi by 3 R Q is
equal to K this is 84 PI into T plus C
the rest of the six pi PI L cancel in
both all three correct so I get R Q as 3
by 4 into a 50 40 plus 1 this is so if
you solve this you get our as nothing
but 63 T plus y because 84 into 4 by 3
is 3 by 4 if you see this becomes 21 21
into 363 this becomes 99 327 this is
what I get R so this is the value of R
at T second 63 t plus 27 to the power 1
by 3 correct please understand this
which is very difficult question you can
say our very talented question if you
understand this question properly here
we have been told that the volume of
this balloon change is constant the rate
of change of volume is constant I wrote
this equation DP by D smoothly
I knew V squared - 4 by 3 PI R I am told
that T is equal to 0 is equal to 3 T is
equal to 3 hours go to 6 and T is equal
to TF / the power that means I have
these equations this equation is nothing
but this equation right if you solve
this and that is the differential
equation so from here onward the
question is simple
let's take one more question the
question says in the bank the principal
increase continuously at the rate of odd
percent per year
find the value of odd if these hundred
doubles itself in ten years
please note this is similar to the last
question the first questions the first
line says that the principal increase
continuously at the rate of our percent
right that means is my P is the
principal then DP by DT this constant
there is our percent of T there is R by
hundred into P correct that is what the
question says now from this we can
easily find the equation because this is
nothing but differential equation if you
see DV by T some K into P right also we
had told that at T is equal to zero my
principal was hundred and at t is equal
to 110 my principle becomes 200 please
not because their B's 100 doubles itself
in 10 years so T is equal to zero it was
hundred rupees and after ten years till
becomes 200 rupees so this is my
question let's solve this so this
becomes Ashley DP by P is equal to R by
hundred DT integrate both side you get
log of P is equal to RT by hundred plus
some constant now to find a value of R
and K using these equations that could T
is equal to zero Pease got two hundred
Spygate
log of hundred is equal to zero plus K
so I got K as log of hundred right so my
K is no same thing
I'll put T is equal to 10 here so this
becomes log of T becomes 100 because P
is 200
this becomes R into T is ten by ten
hundred plus key that is lava hundred so
you get R by 10s log of 200 minus log
100 thank log of 200 minus log of
hundred and that thing well our by 10 is
going to log of two hundred by hundred
nothing with love to and law of two is
0.6931 correct so R is going to turn
into log of 2 that is 0.6931 so R is
nothing but 695 and that is my answer
all right not that difficult to find
easy to find this so what we have seen
here is if I told that rate of change
the principal is constant DV by DT is
some constant and that is R by hundred
in to be right it's not contact ly this
is our banana D into P we had told that
DT by DT is our binary into P because it
it is proportional to P n R so we got
this equation this is a differential
equation and this is what an example of
application of differential equation in
the real life this I didn't tell in the
first line we can use this also and then
the money bank grows in the pan this
goes jelly with this interest money
grows and the population grow everything
grow with the differential equation and
this I was told that T is equal to zero
math with 100 please will retain govern
and I supposed to find the rate of
interest and then this icon the rate of
induce thank you visit exam viacom to
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