JEE Exam > JEE Videos > Mathematics (Maths) Class 12 > Examples : Linear Differential Equations
Examples : Linear Differential Equations Video Lecture | Mathematics (Maths) Class 12 - JEE
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FAQs on Examples : Linear Differential Equations Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. What is a linear differential equation? |  |
Ans. A linear differential equation is a type of differential equation in which the dependent variable and its derivatives appear only as linear terms. It can be written in the form:
a_n(x)y^n + a_{n-1}(x)y^{n-1} + ... + a_1(x)y' + a_0(x)y = f(x),
where y is the dependent variable, y' is its derivative, a_n(x) to a_0(x) are functions of x, and f(x) is a given function.
| 2. How can I solve a linear differential equation? | |
Ans. To solve a linear differential equation, one can use various methods such as the method of integrating factors, variation of parameters, or using the Laplace transform. The specific method to use depends on the equation's form and coefficients. By applying these methods, one can find the general solution or a particular solution to the equation.
| 3. What is the order of a linear differential equation? | |
Ans. The order of a linear differential equation is determined by the highest derivative that appears in the equation. For example, if the highest derivative is y''', then the equation is of third order. The order of the equation affects the number of initial conditions needed to find a unique solution.
| 4. Can a linear differential equation have a non-linear function on the right-hand side? | |
Ans. No, a linear differential equation cannot have a non-linear function on the right-hand side. The right-hand side of a linear differential equation must only contain linear terms involving the dependent variable and its derivatives. If a non-linear function appears, the equation becomes non-linear.
| 5. Are linear differential equations widely used in real-world applications? | |
Ans. Yes, linear differential equations have numerous applications in various fields such as physics, engineering, economics, and biology. They are used to model and analyze various phenomena, including population growth, heat transfer, electrical circuits, and chemical reactions. The linearity property allows for easier mathematical analysis and solution techniques, making linear differential equations valuable in practical applications.
Text Transcript from Video
hello friends this video on differential
equation part 18 is brought you by exam
fear calm no more free from exam before
watching this video please make sure
that you have watched part 1 to part 17
let's take one more question
dy by DX plus 2y it's for a sine X we'll
compare this guy with py is when did you
I get P is equal to Q and using sine X
is very easy to find we do the power
integration of P that is 2 dot DX let us
meet the god place I got I F so my
equation is y 2i f is equal to Q into I
F integration dot DX plus C or Y into e
to the power DX is equal to integration
of Q is sine X integral a to the power 2
X plus C that is what I get now the
trick here is to find the integration of
this so if you find the integration of
this which you have to spend out of time
on that you can use the you can refer my
videos on integrals where I have found
the intuition of all such kind of
questions then you'll get this value as
I'll just write the value I want so for
you e to power v du sine X minus process
this is a tricky part you have the use
of integration formulas to solve this
you get this as well correct so my
solution is still seen so I get Y as my
son thanks Lance OS X plus C see it is
bar minus so here in this question is
it'll any part so I'm not solving this
because in this chapter we are not
concerned about learning how to
integrate sine x integral part of X that
thing we have covered in the
exactly here we never have more
interested in finding the solution for
this differential equations let's take
one more example all square x dy by DX
plus y is it going to tan X so this
equation I can write in this form dy by
DX plus y by cos 4x and X 1
Pascual X right so my P is 1 by cos 4x
that is 6 4 SP is 6 square legs below P
and Q is tan X by cos square X so I can
say 6 6 4 X tan X is all of me in you
same thing had to here
I also find I have first e to the power
integration of P dot DX that is e to the
power in addition of Seck square X DX
that is V to the power x this is my ia
my solution is y into I F is equal to
integration of Q what I have dot T X
plus some constant so it you get Y into
e to the power tan and X is equal to
integration of Q is what sects per X 10
X into y f is e to the power tan X right
DX plus C so here you can take e to the
power tan X as T Atlee
so this becomes e to the RT T into 1
minus T Square actually so if you solve
this guy if you solve this guy what
you'll get is can X minus 1 e to the
power tan X this is what you do so not
that difficult actually you can just put
an X infinity e right so you get 6 for X
becomes 1 minus tan
square one minus T Square into T into e
to the power T sorry this is anyway corn
actually DT actually so you got T e to
the power T only right so you can solve
this very easily so not a big not
difficult integration so you get this
value now to find y I get Y is equal to
tan X minus one plus C into e to the
power minus 10x and there is matter
nothing great here we just found that
this mission is of the form first-order
linear differential equation use the
formula which if you write for the I F
the solution is y into I have is equal
to integration of Qi FD X plus C let's
take up example X divided by DX plus 2y
is equal to X square log X you
rearranged you get dy by DX plus 2 by X
into y is equal to X log X this because
P and this for the incomes Q type so P
is 2 by X and Q is X now X finding I is
not a difficult he to the power
integration of P dot DX that is e to the
power integration of 2 by X DX that is e
to the power 2 log X that is e to the
power log X square and that is X square
so I it is x squared spallation is y
into ia is equal to Q into our
integration dot d expression so Y into X
square is equal to integration of Q that
is X log X into I F then is X square DX
plus C correct so now we have to find
the integration of this that is log X
into X cube right so if you solve this I
won't solve this for you can solve this
this comes out to be log X integration
followed by four minus x over by 16 plus
see that's what you get correct
so this is why X square because this
whole thing is this banners so now Y or
this is my answer as you can see or you
can see that Y into X square is going to
1 by 16 if you take common this becomes
log of X for log of X into HT power
porosity common minus 1 plus that is my
stake my move 1 plus X square dy plus 2
XY DX little cot X DX rearrange
everything so if you rearrange
everything what you get is dy by DX plus
2xy by 1 plus X square is X you divide
everything by 1 plus x squared so this
guy becomes P is the against Q that is P
is UX by 1 plus x squared and Q is
caught XY correct this Cod so I got the
value of P n Q integrating factor I can
find is nothing but e to the power P dot
DX that is e to the power integration of
DX by 1 plus X square DX that is if you
see into the power log of 1 plus X and
this guy is nothing but 1 plus s so I
got the Rollo I F my solution is y into
I F is equal to integration of q dot i m
dot d X plus some constant that is y
into 1
X square because I F is 1 plus X square
integration of Q that is caught of X by
1 plus X square into I F that is 1 plus
X square DX is canceled so integration
of cortex is nothing but log of sine X
this is log of mod of sine X plus X so
that is my solution Y into 1 plus X sine
X plus noise all right so not that
difficult such a complex equation but we
just solve it using the approach one
more question X plus 3y squared dy by DX
is equal to Y so restructure this you
get dy by DX is equal to Y by X plus C
if you see this is not a homogeneous
differential equation so we can't use
that so let's break this so if you see D
by DX is this form right it's difficult
for me to form dy by DX 1 if I flip it I
can easily get the answer so let's see
that if you say DX by dy this becomes X
by y plus dy square correct or I get DX
by dy minus X by Y is equal to 3y so in
this case my P will be 1 by a by 2 minus
1 by Y and Q will be 3y
so my I F will be e to the power P dot
dy here because we have e to the power
minus 1 by Y dy
the base is minus log point that is that
is e to the power log of 1 by Y and that
is 1 by 1
please no 1 e to the power minus log Y
is not - right is 1 by Y is a critical
step here e to the power minus log Y is
1 by Y right so I got ifs 1 by Y so my
solution in this case will be X into
that is DX by D waveform is equal to
integration of Q into I f dot dy
responsible right so this becomes X into
y of there is X 1 by Y is equal to
integration of q3y into I F is again 1
by Y dy plus C it's cancelled so this
becomes 3 y plus C so I get X as 3 y
squared LC y that is my solution general
solution for this should not difficult
the only thing was here we got dy by DX
as this was difficult to solve I just
flipped it right this step is critical
here so it got DX by device but I got
this form DS by D Y as well do P of X
plus D of D DX by dy plus P of X will be
Q so I got P and I got Q thank God I am
and one more step here is to take it is
e to the power minus log Y is 1 by Y
naught naught minus y o because if the
power minus log Y is e to power law of 1
by way thank you visit exam viacom to
watch free it should be should be doers
try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thanks and sorry
equation part 18 is brought you by exam
fear calm no more free from exam before
watching this video please make sure
that you have watched part 1 to part 17
let's take one more question
dy by DX plus 2y it's for a sine X we'll
compare this guy with py is when did you
I get P is equal to Q and using sine X
is very easy to find we do the power
integration of P that is 2 dot DX let us
meet the god place I got I F so my
equation is y 2i f is equal to Q into I
F integration dot DX plus C or Y into e
to the power DX is equal to integration
of Q is sine X integral a to the power 2
X plus C that is what I get now the
trick here is to find the integration of
this so if you find the integration of
this which you have to spend out of time
on that you can use the you can refer my
videos on integrals where I have found
the intuition of all such kind of
questions then you'll get this value as
I'll just write the value I want so for
you e to power v du sine X minus process
this is a tricky part you have the use
of integration formulas to solve this
you get this as well correct so my
solution is still seen so I get Y as my
son thanks Lance OS X plus C see it is
bar minus so here in this question is
it'll any part so I'm not solving this
because in this chapter we are not
concerned about learning how to
integrate sine x integral part of X that
thing we have covered in the
exactly here we never have more
interested in finding the solution for
this differential equations let's take
one more example all square x dy by DX
plus y is it going to tan X so this
equation I can write in this form dy by
DX plus y by cos 4x and X 1
Pascual X right so my P is 1 by cos 4x
that is 6 4 SP is 6 square legs below P
and Q is tan X by cos square X so I can
say 6 6 4 X tan X is all of me in you
same thing had to here
I also find I have first e to the power
integration of P dot DX that is e to the
power in addition of Seck square X DX
that is V to the power x this is my ia
my solution is y into I F is equal to
integration of Q what I have dot T X
plus some constant so it you get Y into
e to the power tan and X is equal to
integration of Q is what sects per X 10
X into y f is e to the power tan X right
DX plus C so here you can take e to the
power tan X as T Atlee
so this becomes e to the RT T into 1
minus T Square actually so if you solve
this guy if you solve this guy what
you'll get is can X minus 1 e to the
power tan X this is what you do so not
that difficult actually you can just put
an X infinity e right so you get 6 for X
becomes 1 minus tan
square one minus T Square into T into e
to the power T sorry this is anyway corn
actually DT actually so you got T e to
the power T only right so you can solve
this very easily so not a big not
difficult integration so you get this
value now to find y I get Y is equal to
tan X minus one plus C into e to the
power minus 10x and there is matter
nothing great here we just found that
this mission is of the form first-order
linear differential equation use the
formula which if you write for the I F
the solution is y into I have is equal
to integration of Qi FD X plus C let's
take up example X divided by DX plus 2y
is equal to X square log X you
rearranged you get dy by DX plus 2 by X
into y is equal to X log X this because
P and this for the incomes Q type so P
is 2 by X and Q is X now X finding I is
not a difficult he to the power
integration of P dot DX that is e to the
power integration of 2 by X DX that is e
to the power 2 log X that is e to the
power log X square and that is X square
so I it is x squared spallation is y
into ia is equal to Q into our
integration dot d expression so Y into X
square is equal to integration of Q that
is X log X into I F then is X square DX
plus C correct so now we have to find
the integration of this that is log X
into X cube right so if you solve this I
won't solve this for you can solve this
this comes out to be log X integration
followed by four minus x over by 16 plus
see that's what you get correct
so this is why X square because this
whole thing is this banners so now Y or
this is my answer as you can see or you
can see that Y into X square is going to
1 by 16 if you take common this becomes
log of X for log of X into HT power
porosity common minus 1 plus that is my
stake my move 1 plus X square dy plus 2
XY DX little cot X DX rearrange
everything so if you rearrange
everything what you get is dy by DX plus
2xy by 1 plus X square is X you divide
everything by 1 plus x squared so this
guy becomes P is the against Q that is P
is UX by 1 plus x squared and Q is
caught XY correct this Cod so I got the
value of P n Q integrating factor I can
find is nothing but e to the power P dot
DX that is e to the power integration of
DX by 1 plus X square DX that is if you
see into the power log of 1 plus X and
this guy is nothing but 1 plus s so I
got the Rollo I F my solution is y into
I F is equal to integration of q dot i m
dot d X plus some constant that is y
into 1
X square because I F is 1 plus X square
integration of Q that is caught of X by
1 plus X square into I F that is 1 plus
X square DX is canceled so integration
of cortex is nothing but log of sine X
this is log of mod of sine X plus X so
that is my solution Y into 1 plus X sine
X plus noise all right so not that
difficult such a complex equation but we
just solve it using the approach one
more question X plus 3y squared dy by DX
is equal to Y so restructure this you
get dy by DX is equal to Y by X plus C
if you see this is not a homogeneous
differential equation so we can't use
that so let's break this so if you see D
by DX is this form right it's difficult
for me to form dy by DX 1 if I flip it I
can easily get the answer so let's see
that if you say DX by dy this becomes X
by y plus dy square correct or I get DX
by dy minus X by Y is equal to 3y so in
this case my P will be 1 by a by 2 minus
1 by Y and Q will be 3y
so my I F will be e to the power P dot
dy here because we have e to the power
minus 1 by Y dy
the base is minus log point that is that
is e to the power log of 1 by Y and that
is 1 by 1
please no 1 e to the power minus log Y
is not - right is 1 by Y is a critical
step here e to the power minus log Y is
1 by Y right so I got ifs 1 by Y so my
solution in this case will be X into
that is DX by D waveform is equal to
integration of Q into I f dot dy
responsible right so this becomes X into
y of there is X 1 by Y is equal to
integration of q3y into I F is again 1
by Y dy plus C it's cancelled so this
becomes 3 y plus C so I get X as 3 y
squared LC y that is my solution general
solution for this should not difficult
the only thing was here we got dy by DX
as this was difficult to solve I just
flipped it right this step is critical
here so it got DX by device but I got
this form DS by D Y as well do P of X
plus D of D DX by dy plus P of X will be
Q so I got P and I got Q thank God I am
and one more step here is to take it is
e to the power minus log Y is 1 by Y
naught naught minus y o because if the
power minus log Y is e to power law of 1
by way thank you visit exam viacom to
watch free it should be should be doers
try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thanks and sorry
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About this Video
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Nov 22, 2024 Last updated |
Video Description: Examples : Linear Differential Equations for JEE 2024 is part of Mathematics (Maths) Class 12 preparation.
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