Examples : Homogeneous Differential Equations and their Solution Video Lecture | Mathematics (Maths) Class 12 - JEE

hello friends this video on differential
equation part 15 is brought to you by
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before watching this video please make
sure that you have watched part 1 to
part 14 so let's take one my example dy
by DX is equal to X plus y by X so here
this guy also homogeneous differential
equation why because if you see this
there is a power 0 so to confirm whether
this is a homogeneous equation will not
against you this f of XY is this guy X
plus y by X let's try to find f of
lambda x and y this becomes lambda X
plus lambda y by lambda X lambda will be
common this becomes X plus y by X
correct this is what you get so f of
lambda X Y is f of X that means it is a
homogeneous differential equation so my
equation of this form dy by DX equal to
this right so I can say Y is equal to VX
so if you say Y is equal to VX you get
dy by DX as V Plus X DV by DX right and
this guy is equal to X plus y by X Y is
equal to VX so I can write DX x plus DX
by X that is 1 plus B so if we solve
this father does 3 and we gets cancelled
what you get is or DV integration there
is nothing but DX by X correctly so what
you get is logs X is equal to B plus C
naught since you have got the solution
let's replace B by Y by X so this
becomes log X is equal to
x+ and that is my answer very simple the
first thing was fine exactly if it is a
homogeneous differential equation or not
if yes
replace Y is equal to VX and then get
the answer but more similar question X
minus y dy is equal to X plus y DX if
you rewrite this equation this becomes
dy by DX this is equal to X plus y MX
now is this a differential equation yes
because I can write this guy as I
suppose X into 1 minus y by X Y 1 plus 0
it yeah you can write it in this fashion
that is nothing but G of my breaks right
so this is the differential equation
homogeneous differential equation in
this case I can use Y is equal to VX
deficient both side you get are dy by DX
is equal to V Plus X into T V by DX and
dy by DX is nothing but we know is there
one class Y by X Y by X is V actually if
you see 1 plus V by mothers America is
correct or you add X DV by DX is 1 plus
V by 1 minus P minus P correct and this
is nothing but 1 minus V 1 plus V minus
V Plus v square correct so you get 1
plus v square very balanced so this
photo this becomes DV into 1 minus V by
1 plus v square is equal to DX by X
correct so now we have to integrate both
side
so we have to really write this at
length so this I can write as 1 by 1
plus V Square D V minus V by 1 plus V
Square D V correctly is equal to DX by X
you solve this guy this becomes tan
inverse V minus 1 by 2 log of 1 plus v
square is equal to X plus some long
expression was corrected and then you
get to place V with Y by X what you get
is tan inverse Y by X minus 1 plus y
square by X square is equal X plus in
that is my cut let's take one example X
square dy by DX is equal to X square
plus 2 y square minus so I can write dy
by DX as X square minus 2 y square plus
XY by X square now let's check if it is
a homogeneous or not this I can write as
you take X square common 1 minus y
square by X square plus y by X so if you
see that it is nothing but sum function
of home G by by X that means it is a
homogeneous equation so I can say that Y
is going to be X again replace so this
becomes dy by DX as V Plus X into DP by
DX correct I'd dy by D is equal to this
whole thing X square sorry this guy 1
minus 2y y by X is V 3 square plus V
correct VB cancel
so what I get DV by 1 minus 2 e square
is equal to DX by X correct if we
integrate both sign so in this case a
square minus X square for my integration
so I can take 1 by 2 common here so this
becomes 1 by 2 correct so this becomes 1
by root 2 actually whole square so if
you solve this this way everything but 1
by 2 in do love my room plus y by X this
be 1 by root 2 minus weakness bye-bye X
is equal to log of mod X plus C so if
you solve this further this becomes this
becomes root 2 and this becomes 1 by 2
root 2 log of this becomes X plus 2 y
this becomes X minus y is equal to log
of mod X plus root 2 root 2 cancel thank
you so that is what you get so we're
able to find solution very easily first
thing we did was we found that this
guy's of Malini's differential equation
and then we use Y is equal to VX let's
take some more example X dy minus y DX
is equal to root of X square plus y
square I want to find dy by DX in this
case my dy by DX in this case will come
to be go of X square plus y square plus
y all the DX component by dy combined it
is X so if you see I can write this
guy's nothing but X common root of 1
plus y square by X square plus y by XY
it's that is geo of some WebEx that
means I can say that this guy's a linear
differential equation since these guys
are homogeneous the relocation I can use
Y is going to be X so I get dy by DX
plus X DV by DX and dy by D is this guy
and that is root of 1 plus by x is v v
square plus v correct this and this is
cancel so what i get DV by root 1 plus v
square into DX max you integrate both
side you get log of V plus root 1 plus v
square is equal to log of mod X plus C
so this is a formula we have used if you
don't understand this step please watch
my integral videos where I explain how
do u by root 1 plus v square is going to
log of V Plus
since everything is in Log form let me
take logs here so what you get is log of
V Plus V is I can say is my bags writing
input here Y by X plus 2 plus 1 by X is
equal to log of X me i can remove log
log both side so what you get is y my X
plus root X square plus y square PI X
becomes because root of x square becomes
X is equal to XC correct or you get y
plus take square plus y square is not a
very difficult version just first thing
we did was confirm whether it's a
homogeneous differential equation if yes
we use Y is going to be nice and then
just solve the equations thank you visit
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