Applications of Laplace Transforms

Añadido: 2026-05-05

[00:00:10]Good morning all. I am professor swami from SR University department of mathematics.

[00:00:18]Today we are going to discuss the applications of laplas transform. In previous lectures we have discussed the different uh forms of laplas transform right. Laplas transform by differential equations.

[00:00:35]Laplas transform uh differentiation integrals and laplas transform division by s multiplication by s. Right? So we have discussed elaborately in 1 to 7 lectures. Now in this lecture the applications of laplas transform that is to the ordinary differential equation or the solutions of ordinary differential equation by laplas transform method. So how you solve using laplas transform the ordinary differential equation.

[00:01:13]First of all what is the solution of ordinary differential equation? Suppose for example the differential equation is like this. d² y by dx² + 2 dy by dx + 4 y = some e power x. So solve.

[00:01:36]So how you solve normally the differential equations? First write in operator form d² + 2 d 4 d or 2d 2d + 4 into y = first zero.

[00:01:54]Right? What is d here? Capital d is equal d by dx differential operator.

[00:02:00]Next write auxilary equation.

[00:02:03]So what is the auxilary equation? This is nothing but algebraic equation m² + 2 m + 4 = 0. Now solve for roots then we will get root m= plus or minus something.

[00:02:19]Then write complimentary function y is equal.

[00:02:24]So this is a complimentary function for the homogeneous differential equations.

[00:02:30]If equal to zero homogeneous suppose equal to e power x for e x you can you you should calculate particular integral that e power x there are different cases to finding a particular integral. There are many cases according to that cases find a particular integral. Then the finally the general solution of differential equation is y is equal complimentary function cf + pi particular integral.

[00:03:04]This is the general solution of the given differential equation ordinary differential equation.

[00:03:10]Right?

[00:03:12]But if you apply the laplas transform to the ordinary differential equation, no need to calculation of this complimentary function in particular integrals, right? And adding of complimentary function particular integral. Yes. To apply the laplas transform at a given initial conditions it will give direct solution. No need to calculate this complimentary function particular integral. How? As you see here that is the advantage of laplas transform right the laplas transform can be used to solve the differential equation with constant coefficients only. The advantage of laplas transform is particular solution can be obtained for the given initial conditions without obtaining the general solution that is a advantage.

[00:04:05]So what is the working procedure here?

[00:04:07]The working procedure is first write down the given differential equation right and apply laplas transform on both sides to the given differential equation and use the given initial conditions.

[00:04:24]Substitute the initial conditions and rearrange the equations.

[00:04:30]Right? And rearrange mean just simplify for the solution. Next finally you should take inverse laplas trans on both sides then it will give the solution.

[00:04:46]First write the given equation in proper way and apply laplas transform both sides and apply substitute initial conditions and simplify and finally apply inverse laplas.

[00:04:59]These are steps to find the solution of ordinary differential equation using laplas transforms.

[00:05:10]Just you go to this. You know the laplas transform derivatives already we discussed in previous lectures. The laplas transform first derivative is s intolus f0 or you may write this s into laplas f is f of s or right f of s f of0. Similarly the second order derivative is s² into fr of s minus s into f of0 minus fdash of0 second order d.

[00:05:47]Why we are introducing here laplas trans derivatives? This is for this is required for the solve the ordinary differential equation. Laplas trans derivatives required here. You see here is a third order derivatives. These formulas are now useful to solve the ordinal differential. The fourth derivative. Third derivative start with sq bar of h and s power decreases s² s and s^0 here. Now here f of s² f of0 fdash of0 fdash of0 like that. So here fdash end with f dash of f4 derative end with f3 f with nus like that here n s n -1 n - 2 n - 3 and so on.

[00:06:39]This is the LLA transform derivatives.

[00:06:42]Now let's go to the example how you solve the differential equation using laplas transform. Solve the following differential equation by laplas transform method. Here d² x by dt² - 4 dx by dt dt minus 12 x = e ^ of t where the initial conditions x of0 x of0 means x of t=0 here is 1 x dash t=0 is -2. Now this differential equation you can write this x dash t - 4 derivative of t - 12x of t = e^ so this is the given differential equation right this this can be write in the form of either x is a dependent variable and t is independent variable right both side taking inverse applause right sorry not inverse applause First both side take laplas transform not inverse laplas first laplas finally inverse laplas. So laplas transform of x of t is xar of s. Laplas transform of x dash t is s into xar of s minus x of0.

[00:08:11]Right? Already we discussed in previous slide. Now laplas transform x dash t is s² x of s into x of0 minus xdash of 0. This is a second order derative. Now here you apply this transform of this x dash is this one x dash t. Next x of t next x of laplas trans of this one. Laplas trans of x of is xar of s. x dash t is this x dash is this. Just substitute and the initial conditions x of0 is 1 apply and x0 is -2 just you write down here and simplify here s s² x of s - x right - s + 2 4 x - here initial condition is 1 and s by 1 1 by sus 3 Right? Laplas transform this e to the power 3. Now this is xar of s take out xar of s take out. So s² - 4 s - 12 your remaining terms minus s right + 6. Where is 6 here? - 4 into -1 + 4 here + 2 + 6 = 1x s - 3. Is it right? Now xar of s s² - 4 s - 2 side - s + 6 to right side. So plus s here - 6 by here 1 by sus 3. Now next is solve for roots right sir what are the roots here s² - 4 s - 12 =0 what are the roots here s² right uh 12 6 2 are 12 right or 3 4 are 12 is 4 now here what do you write here uh plus here - 4 s 6 6 Yes s + 2 s - 2 l = 0 is it right? - 6 + 2 is -4 - 6 into + 2 is -2. So s - 6 here + 2 common s - 6 is = 0. So s - 6 and s + 2 are roots. s - 6 s + 2 are roots. Now xar of s= this side. This is xar of s right. Now both side taking inverse laplas.

[00:11:23]Take inverse laplas on both sides. So inverse laplas x bar of s and inverse laplas of this.

[00:11:32]So how you solve this inverse 1x s + 2 s - 3 s - 6 plus directly you you know e ^ - 2t form inverse s + 2 e^ - 2 but this is you can't get how you get the inverse laplas on this type of function speaking to partial fractions speaking to partial fractions a by is placed to b - 3 c by s - 6.

[00:12:07]Now take LCM and solve and put a = 3 and a c = 6 and obtain a b c values.

[00:12:19]Then we get a = 1x 40, b = -1x 15, c= 1x 20. Already we did many problems on splitting into partial fractions. The similar method you apply and obtain ABC values and substitute here right 1x 40 L inverse of 1x s + 2 B is - 1 by 15 and C is S1 right 24 like this.

[00:12:49]Now the inverse laplas transform of 1x s + 2 is e ^ -2 1x s - 3 e ^ 3t 1 by 24 s - 6 e ^ of 6 your 1x s + 2 e ^ - 2t so e ^ - 2t here also e ^ - 2t take out here so 1x 40 + 1 is 40 1 by 40 e ^ - 2t - 1x 15 right e ^ of 3t 1x 24 e ^ of 60 this is a solution of the given differential equation by applying inverse or laplas trans not only inverse applying transforms this is a best method to solve the ordinal differential equations right and next example let's take this example Here y dash - 3 ydash + 2 y = 4t + e ^ of t where y of 0 is 1 yd of 0 is -1 here y is a function of t now y is the function of t right because is a y depends on t now 4 t time domain so this are second order apply laplas transform for second order derative right and first order derative and this is zero order and like that so inver sorry laplas transform y dash t minus 3 laplas transform y- of t plus 2l of tapl both sides apply laplas transform you know the formula y dl is this s²us s² y fs S into Y of 0 - Y of 0 - 3 into Y is S into Y of 0. Right? This the formula first derivative formula second order derivative formula already we discussed in previous slide and Y of TL Y bar of S and LLA transform of T. What is the transform of T? T 1 factorial by S^ n + 1. So s square one factor s² right is it right is 1x s² already four is there so 4 y e^ 3 + 1x - 3 now apply this initial conditions y of 0 is 1 yd of 0 is -1 just you put y of 0 is 1 yd of 0 is -1 y of 0 is 1 y bar of s. Now simplify this.

[00:15:50]Take out y bar of s common and simplify y bar of s is right s² - 3 s + 2 here min - s + 4 the remaining terms and send to right hand side right and these roots of this s² - 3 s + 2 =0 what are the roots s² here - 2 s - s - 2 into -1 is + 2 - 2 -1 is -3 + 2 = the are roots. So s common s - 2 here -1 common s - 2 here - so roots are sus1 and s -2 these are roots right now s -2 and s -1 to right hand side. So 4x s² into this and 1x s -1 s - 20 this and s - 4 by this. So if you take the LCM LCM of this function is s² s -1 s - 2 s - 3 and this is a numerator and this is a denominator right. So y bar of s is equal right s^ 4 - 7 sq + 13 s² + 4 s - 2 l by s1 s² right here s 2 1 3 4 5 s^ final so split into partial fractions inverse laplas transform split into partial fractions right this is a big fraction so a by S d by s² c by s -1 d by s - 2 e by s - 3 is it right? So now the simplifying and uh calculate a value is three after calculation b value is 2 c value is - 1 by 2 d value is -2 e value 1 by 2 then directly substitute here a b c values so three inverse laplas transform here 1x s apply inverse laplas on both sides 1x s 1x s is Laplas transform constant is constant by S. So 1 and inverse transform 1x s² is t now and inverse 1x s - 1 e ^ t 1x s - 2 e ^ 2t 1x s - 3 e ^ 3t. This is the solution of given differential equation by applying laplas transforms. Right? And here one more problem is a d² x by dt² + 9x = sin t sin t here x of0 is 1 x dash of 0 is -1 is it right?

[00:19:03]Similar procedure just to follow. Take inverse applications from both sides.

[00:19:08]Here second order derative. Here first order derative to no first order simply only x of x bar of only second order. Next x. So laplas of sin is 1 by 1 + s².

[00:19:26]Right? Here the conditions of x of0 using initial conditions x of0 is 1.

[00:19:33]Here x of<unk> by2 is -1 and x dash of 0 is not given in the given function x dash of zero is a right here you observe here only two conditions x of0 is 1 right and x dash of 0 is minus1 right but here x dash of0 is not given this is x of this is not given this is actually finally we get just you assume x of 2 is -1 by 2 right so xdash of0 is assume a x dash of 0 is assume a so finally you would get a value also so by substituting this x of0 is 1 x dash of0 is a x dash of 0 is a Right.

[00:20:34]Now just you take this simplify this then agree and s² + 9 this one and send to this functions.

[00:20:46]So s by s² + 9 a by send this side taking inverse laplas transform both sides.

[00:20:54]So inverse laplas runs on this uh s split into partial fractions you would get this 1x s² + 1 s² + 9 is equal you can write this into partial fraction and this is directly you know this is cos 3t this is sin 3t right substitute in equation number one so Then 1x8 inverse laplas on this 1x 8 inverse laplas on this like this. This is a already. Now this is sin t. This is - 1x 8. This is sin 3t by 3. This is cos 3t and this is uh sin 3t by 3. So a by 3.

[00:21:44]Now sin 1x 8 here. Here sin 3t and sin 3t take out here right taking t = 5 by2 that is x of 2 is -1 it is given x of 2 is -1 now simplifying this this conditions so you will get a = 7 by 2 right finally we get this solution and next is third order. Previously we discussed only first order and second order differential equation. Now this is for third order differential equation. dq dq + or - 2d² sorry + 5 =0. So you know third order this is a operator form just you write down y dash - 2 y dash + 5 y dash = 0 d is the d is equal d by dx now right so y dash means yle dash means dq y by dxq so then apply laplas transform both sides third order derative this is second order derivative and first order derivative equal to zero. Here homogeneous equation equal to zero. Right? Now simplify this.

[00:23:16]Use these conditions y of initial conditions. So y- of0 y dash of 0. Use initial conditions and solve this. So now applying initial conditions you would get this y bar of s take out here s cube here right - 2 s² + 5 s -1 this is - 1 =0 so y bar of s is 1x sq - 2 s² + 5 plus 5 yes here 5 yes so now s take out 1x s² - 2 s + 5. Is it right? Now y bar of s = 1x s into s² - 2 s + 5. So how you find the inverse laplas of this? So let's take this s² - 2 s + 5 s² - 2 s + 5 = 0. We can't get roots. So then what we get is just rearrange this s1 s - 1 square + 2² is it satisfy this? Yes. S² - 2 s + b² 1 1 + 2² is 4. So five satisfy. So write down s 1x s - 1 square + 2² into s right. Apply inverse laplas transform both sides. So 1x s - 1 square + 2² this is equal e ^ of t by 2 sin t how how it is possible the inverse laplas transform of f bar of s bar of s by s division by s formula division by s formula 1x s division by s right division by s is integral 0 to2 f of t.

[00:25:19]So this is equal this is f bar of s.

[00:25:24]This is f of s. So from this calcate f of t. F bar of this is this f of t is e to e ^ of t sin tx2. So integrate 0 to t e ^ of t sin tx2 dt. Here it is in the form of e power a x sin bx formula integration of e power ax. So what is that? e power a x by a² + b² into a sin bx - b cos bx. So here e a x by a square b a sin bx. Here a = 1 b = a sin bx minus b cos bx and integral 0 to 2. This is a formula right and substitute limits.

[00:26:17]The upper limit same as this function.

[00:26:19]Lower limit this is e^0 is 1 sin 0 is 0.

[00:26:23]Here cos 0 is 1. So minus of minus 2. So this is equal what do we get here?

[00:26:34]Just simplify this 1x 5 5x2 e^ cos 2t here 1x 10 e ^ of t sin that's all this is the given difference you know already the trans bar of s by s is integral 0 to right this is about uh solution of the given differential equation similarly we can solve the uh differential equations by using laplas transform method. So here you may take this d² y by dx² dy by dx + 5 y = e ^ of - sin t y of 0 yd of 0 = -1.

[00:27:23]Similar procedure here also apply both side laplas write in the laplas transform and substitute initial conditions and simplify you would get the solution.

[00:27:36]This is also second order and this is also second order x d² + 1 into x right x means here this is uh d² x by dt² is nothing but function of t plus x of t = t cos 2t t cos 2t how you get the laplas trans 2t so multiply Integration with T cos A T sin A already we discuss T cos A T sin A. So A is equal to here just you write A is equal to so multiplication with T formula apply. So we get this terus t sin t e power minus t sin t apply first shifting theorem first shifting theorem right what is the first shift already we discussed right so apply first shifting theorem and solve this and this is multiplication method and next is d² x by dt² - 2 dx by dt + 2x of t =0 here X this is X dx by dt is = 1 at t =0 means simply xdash of zero is = 1 x of right x of 0 is = 1 this is x of0 and x dash of 0 is only one combined conditions I have given and x of0 x dash of0 initial condition then solve the differential equation. This is the advantage of solution of uh ordinary differential equation by applying the laplas transform and inverse laplas transform. Using this inverse laplas transform llas transform method can solve highly order differential equations also highly complex differential equations also. Sometimes the differential equations not solved by the ordinary methods. What is ordinary method? Complimentary function place particular integral. It maybe fails then go to the laplas trans and solve the uh differential equation applying laplas trans method. Thank you very much.