Solving first order differential equation using Laplace transform

in this video we're going to try and continue working on this uh first order differential equations we will use one more technique of this lapas transform to make it clear and easier here we are trying to solve this differential equations y Prime + y is e to- x with this initial condition y of 0 is 1 we want to use laas transform to find this solution here and we have this hint here to compute La plus of x e to minus X so uh this means that we going to need La plus transform of this x e to minus X okay and uh uh to compute this one this laas of uh x e to minus X we use the definition and we try to use uh any integration technique most of the time here because how we have this x we're going to use integration by part so you can try and use it okay so let's solve this one to solve this one we're going to apply laas transform on all the sides the left and right side so we have La plus of Y Prime + y is equal to La plus of e to- X okay and uh uh remember that laas of e to ax is 1/ s to minus a with s bigger than a okay the sign switches here so that means this one here taking a as minus one okay you need to know how to do this ones you have to look carefully here so here if we have a is minus one we can use it here and get that laas is s + one because we have a is min - - one okay if you are not convinced you can use the definition and get this result exactly on the left side we have this uh sum okay and we know that laas is linear okay so we have laas of Y prime plus laas of y equal to 1 / s + 1 we know that La plus of Y Prime is s La plus of Yus y of Z plus plus of Y so this one here is coming from La plus trans form of Y Prime okay how to get it we can use the definition see one of the videos that we have done and get the result we will have to use integration by part okay now we we collect like terms so here we have S + 1 L of Y is equal to 1 / s + 1 + y of 0 which is 1 so we have 1 / s + 1 + 1 which is going to give us s + 2 / s + 1 and therefore we have lapas of Y is s + 2 / X s + 1 squ okay now we have to use de composition okay fractional decomposition to get the result that we need so we can decompose this one as Alpha over s + 1 + beta over s + 1 s why did we do it this way because these are the rules and you need to learn these ones here we might come back to this okay Crossing cross multiplying or we can just multiply here top and bottom by S + 1 we get alpha s + Alpha over s + 1 s okay and now we can get that this is just alpha s + Alpha + beta over s + 1 s now we can use the uh coefficient identifications okay so in front of s there is one and here in front of s there is Alpha that means Alpha is 1 and Alpha plus beta is two that means Alpha is 1 and beta is 1 so here we solve for beta we get 2 - Alpha which is 1 okay and therefore we have 1 / s + 1 + 1/ S uh + 1 squ okay now we know this one we know La plus uh uh this is L plus of e to Min - x remember here okay and this one is L plus of x e to minus X how did we get this one so if you try to work this one here what you will get is you will get to 1 / s + 1 s okay now we have these two here and now when that laas is linear we get L plus of e to- X Plus e x e to- x equal L plus of Y okay and this means that Y is x e to - X Plus e to - x or or or simply 1 + x e to - x and this is the solution of for our uh equation here okay this is our solution here so we have used the uh many techniques okay and using try to use these hints okay given to you or the table that comes with laas transform to get the result we're solving the first order differential equations to learn all the techniques and remind ourselves of what we have to learn and what we need to solve these differential equations okay I think that's what we need for now and we found our solution and if you see any errors or mistakes please comment also share and subscribe and thank you very much for for everything

Channel: Archimedes Notes

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Id: c6EhKSxyzUc

Length: 7min 2sec (422 seconds)

Published: Sun Jul 14 2024