so welcome to the series of lecture series on mathematical methods and its applications
so on the last lecture we have done some applications of laplace transforms we have
seen that how we can solve an ordinary equation ordinary differential equation using laplace
transforms we have also solved some circuits problem and solved them using laplace
transforms ok now again we will see some more application laplace transform in this lecture
now suppose we have the first problem such problems are basically called integral
equations where where y is unknown function involved under sign of integral so in in these
problems y is unknown function and it involve under sign of integral so such equations are
called integral equations so laplace transform is also useful to solve integral equations
so to illustrate this let us discuss this examples
so first problem first problem is y dash minus four y plus three times integral zero
to t y tau d tau and equal to one equal to t and it is given to you that y zero is so
how can we solve this equation using laplace transform so this is an integral equation
because y is an unknown function involved in under sign of integral also ok so how can
we solve this take laplace transform both the sides so laplace transform of y dash is
p f p p f p minus f zero or y zero minus four times f p plus three times now integral
zero to two y t y tau d tau the laplace of this is nothing but f p by p we already
know this so this is f p y p which is equal to the laplace of p is nothing but one by
p square so here i am calling laplace of this y t as f p ok now it is f p you can take
common f p coefficient f p will be p from here minus four from here and plus three upon
p from here y zero is one given to us so you can substitute y zero as one here so it will
be equal one by p square plus one so this implies f p into p square minus four p plus
three upon p is equals to one plus p square upon p square so p p cancel out this implies
this implies f p is nothing but one plus p square upon upon p into now this is p minus
one p minus three the factor of this so this will be the f p so to find out y t we will
make we will take laplace inverse both the sides
so to find out the laplace inverse of this f p we will make use of partial fractions
so this is nothing but upon p a upon p plus b upon p minus one plus c upon p minus
three so to calculate to calculate a we will substitute p equal to zero in this
expression so when when we take p equal to zero it is nothing but one upon one upon three
so one upon three will be the value of a for b you can substitute p equal to one p
is one it is two upon it is two upon one into minus two that is minus one and when p is
three when p is three it is nine plus one ten upon three into two two fives are ten
it is five by three so these are the values of a b and c ok so what will be y t y t will
be nothing but laplace inverse of f p and which is nothing but laplace inverse of this
expression so laplace inverse of this expression [is] nothing but one by three minus e to the
power t plus five by three e to the power three t so that will be the laplace inverse
of this expression ok so you can check when y zero equal to one
when t equal to zero when t equal to zero this value is one by three minus one that
is minus two by three minus two by three plus five by three is one so this condition
is also satisfied so this is the solution of this integral equation now suppose you
want to solve the second problem is it is again a integral equation so how can
we solve this problem let us see again using laplace transforms this problem is y x is
equal to three x square plus integral zero to x y t sin s minus t d t so this is the
problem ok now how can we solve this integral equation using laplace so again take laplace
both the sides so laplace of y x suppose it is y p suppose laplace y is y p three x square
so laplace x square or t square is nothing but two upon p q plus
now to find out laplace of this expression it is nothing but the the convolution of
y with sin the convolution of y with sin function is this expression so by the convolution theorem
we already know that convolution theorem of laplace transform it is zero to t
f u g t minus u d u is nothing but laplace of this is nothing but f p into g p where
f p is the laplace of function f and g p is a laplace of function g small g so this is
by the convolution theorem of laplace transforms so we again apply a convolution theorem over
here in this expression so we will get laplace transform of y t we have already assumed that
it is y p into laplace transform of sin function is nothing but one upon p square plus one
this is by the convolution theorem of laplace transforms now y p you can collect the
terms containing y p so it is y p into into it is p square upon
p square plus one which is equals to six upon p q now this implies y p is equal to six p
square plus one upon p to the power one so this would be equal to six one by p cube plus
one by p to the power five so to calculate to calculate y t is nothing but laplace inverse
of y p and it is nothing but laplace inverse of this expression so laplace inverse of this
expression is nothing but six into laplace inverse of one by p cube is t square upon
two and laplace inverse of one by t to the power five is nothing but t to the power four
upon factorial four so this is nothing but three t square plus t to the power four upon
four so this is the this is the solution of this integral equation ok
now let us solve next problem again it is a integral equation based on the same
concept ok so in the next problem is that is zero to x y t upon under root x minus
t d t and it is equal to one plus two x minus x square so we have to find that y which satisfy
this expression basically we have to find that y which satisfy this expression so this
is an integral equation basically so how can we solve this again take laplace transforms
both the sides now to find out the laplace transforms of this expression we will again
use convolution theorem for laplace transforms because it is a convolution of under root
x with under root t with y t ok because a because it involve two functions basically
zero to x y t into x minus t to the power minus half so it a con it is a convolution
of this function with function ok convolution of y t and t so we will take the laplace transform
both the sides so laplace transforms of this by the convolution theorem is laplace
transform y t which are which suppose is y p and laplace transform of t to the power
minus half is equal to or f to the power minus half both are same you can take t k power
minus half or s k power minus half now laplace transform of one is one by p two
by p square minus it is laplace transform of t square t laplace transform of t square
is two by p cube so it is y p laplace transform of t to the power minus half is gamma half
upon p to the power half which is one by p plus two by p square minus minus two by p
cube gamma half is under root pi so y p is nothing but one by under root p time one by
under root pi times one by under root pi plus two upon p to the power three by two minus
two upon p to the power five by two ok now y t is nothing but laplace inverse of
this expression so take laplace inverse both the sides so laplace inverse of this will
be nothing but it is one by under root pi now laplace inverse of one by under root p
again recall the formula of t to the power laplace of t to the power n is nothing but
gamma n plus one upon p to the power n plus one so from here laplace inverse of one upon
p to the power n plus one is nothing but gamma is nothing but gamma n plus one upon oh
it is nothing but t to the power n upon gamma n plus one ok
so laplace inverse of this so you substitute n s minus half to get under root p so it will
be t to the power minus half upon gamma half plus two two into one upon p to the power
three by two so you will substitute n s half it is nothing but t to the power half upon
gamma three by two minus again two into it will be t to the power three by two upon
gamma five by two so you can we can simplify this very easily so this will be nothing but
this will be nothing but one by under root pi the first expression is one upon under
root pi into under root t plus two under root t and it is one by two under root pi minus
two into three t to the power three by two upon three by two one by two under root pi
so this we can easily simplify so this after simplification we get one upon pi under
root pi can be taken out common so it is one under root t minus plus four times under
root t minus it is four eight upon three times eight upon three times at t to the power
t to the power three by two so this will be the this will be the final answer ok for this
problem now similarly we can solve the last problem
of this slide again we will take laplace transform both the sides in the left hand
side we have d y by d t so the laplace transform of d y by d t will be p f p minus y zero y
zero given as one and the when the right hand side we will apply convolution theorem for
laplace transforms so this is that is how we can solve integration equations using
laplace transforms ok now next is we can also solve simultaneous ordinary equations using
laplace transforms so let us see some problems based on it so the first problem is
simultaneous differential equation is d x upon d t is equals to two x minus three y
and d y by d t as given as d y d t is y minus two x and given as x x at zero is eight and
y at zero is three y at zero is three so let us solve this problem again again we
will take laplace transform of both the expressions so laplace when we take laplace transform
of first expression so it will be nothing but it is d by d s by d t so it will
be nothing but p into x p minus x zero ok is equals to two x p minus three y p so
that will be laplace of first expression what is x zero x zero is eight so what this expression
implies this implies p minus two times x p ok plus three y p is equal to eight so that
will be the first equation which is obtained here
now apply laplace transform in the second equation when you apply laplace transform
both the sides here we will be having p into y p so this y p is nothing but laplace transform
of y t ok now minus y zero which is equals to y p minus two x p so again this implies
you compare take the coefficients so it is two into x p two times x p and the coefficient
of y p is coefficient of y p is p minus one which is equals to y zero y zero is three
so these are the two equations which we obtain p minus one times y p
so these are two equations which we obtain over here now we will find the these are the
two linear equations in x p and y p ok two linear equations x p and y p so we will solve
them for x p and y p and then take the laplace inverse to find out x t and y t that is a
solution of this simultaneous differential equation so how can we solve this so to find
out the solution it is nothing but hm i mean simultaneous linear equations to into
unknowns so we can solve it like this it is x p upon minus y p upon and it is one upon
so it is the determinant of eight three and three p minus one this is by the cramer rule
we are solving it then minus y p y p coefficient will be eight three c one c two and it is
p minus two and two and here the right hand side will be nothing but p minus one ok p
minus two it is p minus two and three two and p minus one
ok so this implies x p upon it is eight p
minus eight minus nine is minus eight minus nine is minus seventeen minus y p upon it
is sixteen sixteen plus six sixteen plus six is twenty two minus three p it is one upon
it it is nothing but p square p square minus three p plus two minus six ok so this
is x p and y p so what are x p and y p now so x p is nothing but eight p upon [sen/minus]
minus minus seventeen upon it is nothing but p square minus three p minus four which can
be is the factorize it is nothing but p minus four into p plus one which is p square minus
three p minus four of course and y p is nothing but y p is nothing but three p minus twenty
two upon again the same quantity in the denominator now this is nothing but we can make use of
partial fractions again so it is p minus four it is upon minus four it is plus p plus one
so we will substitute p as four first so it is thirty two it is thirty two minus seventeen
thirty two minus seventeen is five fifteen so fifteen upon when you substitute four five
three so it is three this partial fraction you can also compute compare the coefficients
both the sides as you already know now you substitute p as minus one so it is minus eight
minus seventeen upon minus five so it is nothing but minus twenty five upon minus five which
is five so it is five ok now we will use a partial fraction again here
it is minus p minus four plus p plus one so when you substitute p as four it is twelve
minus twenty two upon five so it is minus ten upon five it is minus two and when you
substitute p as minus one it is minus three minus twenty two upon minus five that is minus
twenty five upon minus five it is five ok now to find x t x t which is nothing but laplace
inverse of x p so we will make laplace inverse of this expression three upon p minus four
plus five upon p plus one so it is nothing three e to the power four t plus five e to
the power minus t again to calculate y t we will take laplace inverse of y p which is
nothing but laplace inverse of this expression and it is nothing but minus two e to the power
four t plus five e to the power minus t so these are solutions of this differential equation
you can check also f zero is eight yeah f zero is eight and y zero is three yeah y zero
is three and the it will also satisfy the these equations you can easily verify ok
now let us discuss one more example based on this which is a second order simultaneous
differential equation so this also we can solve using using laplace transforms so
how can we solve let us see so it is nothing but second derivative of x respect to t
which is equals to given as x plus y and d two y by d x two is nothing but it is minus
x minus y ok and the initial conditions are given to you ok
now again take laplace transform both the sides first for the first equation when you
take laplace both the sides so it will be nothing but p square x p minus p into x zero
minus x dash zero ok and it is equal to x p plus y p course x p here is laplace transform
x t and y p is the laplace transform of y t ok now x zero is given to you as two so
this value is two and x dash zero is zero so this implies p square plus minus one times
x p minus y p will be equal to x zero is two so it is two and x dash zero is zero so this
is the first equation which we obtained now apply laplace transform in this equation
both the sides when you take laplace transform both the sides so what we will obtain from
here so when you take laplace transform both the sides here so it is p into y p minus p
into y zero minus y dash zero is equals to minus x p minus y p so this implies now y
zero is minus one given to us and y dash zero is zero so it is x p plus p plus one times
y p y zero is minus one so it is minus it is minus one it is like p so it is minus p
so this is the second equation which we obtained ok so it is it is x p x p will come here
it is plus one times y p and this y zero is y zero is minus one minus minus plus p we
may come here it is minus p now in this again verify this equation this
is f p square minus one time x p and y p x zero is two it is two p ok and minus
y p so this equation is obtained now here it is p square y p p square y p minus p
y zero minus y dash zero equal to this so it is nothing but p square y so what are two
equations which which we obtained it is p square minus one times x p minus y p
is equals to two p and here it is x p plus p square plus one times y p is equals to minus
p so again we will these are simultaneous linear equations so we solve for x p and y
p so it is x p upon a determinant of two p minus p minus one p square plus one is equals
to minus y p upon p determinant of p square minus determinant of two p minus p and
p square minus one upon one and one upon it is determinant of p square minus one minus
one one p square plus one ok so this implies x p upon this expression is
two p q plus two p minus p ok this is minus y p upon this expression is two p minus minus
plus p cube minus p ok and this is one upon p to the power four upon one plus one so clearly
x p will be nothing but it is p cancels it is nothing but two p square plus one
ok upon p cube so it is two upon plus one by p cube and x p y p is nothing but it is
negative of it is negative of one plus one plus p square it is two p minus p is
p p p cancels and it is one plus p square upon p cube so it is nothing but minus one
by p cube minus one by p ok so now x t now x t is nothing but x t we will
laplace inverse of x p and y t is laplace inverse of y t so what will be x t x t
will be nothing but laplace inverse of x p which is equals to two plus now one by p cube
one by p cube is t square by two and y t is will be nothing but minus of two by minus
of t square by two minus one so these are the solutions which we can obtain from here
ok so hence using laplace transforms we ca also solve hm integral equations and
simultaneous ordinary differential equation so in the next lecture we will also see that
how can we use laplace transform to solve partial differential equations so
thank you