Laplace transform to solve first order differential equation.

now we're going to solve the differential equation Y Prime - y = e to the power x with Y of 0 = 1 and note that we have this hint here because we're going to need it so when you have this hint something like this you will have to figure out the n and the a by yourself so be careful this is the general rule here we can prove that uh the lapas transform of x to the^ n e to a x is given by n factorial over s - A to the^ n + 1 here with n bigger than minus one okay now let's use the llas transform to transform this one into a function that we can solve so here using linearity we have L plus of Y Prime - Y is L plus of Y Prime - laplus of Y and it's equal to laplus of e we have proved the that laas of e x is 1 / s- one and using what we know about the laas of the derivative we can get this one here to be S llas of y - y of 0us L of Y = 1 / s -1 okay here we can get this together get S -1 L plus of Y = 1 / s - 1 + y0 and this gives us 1 / s - 1 + 1 and therefore we have laas of Y is uh s/ s -1 squar okay so now we need to decompose this one here and see what we can get so the the the first thing that we're going to do we're going to decompose this one here we can write it as uh uh if we want to uh uh be precise as uh in the following manner so note here that in this case we can write this one as um Alpha over s -1 + beta/ S -1 s okay and if we multiply here by S -1 we get uh this simplification and therefore we have alpha s + beta - Alpha over s -1 s so here here using identification we get that here so Alpha is one and beta minus Alpha is zero therefore we get uh that Alpha is 1 and beta is 1 okay and this gives us that uh L of Y which is equal to 1 / s - 1 + 1 / s - 1 to the power squar now we can uh uh use this fact here okay so here this one okay this is uh this is telling us that a is one okay a is one for the first one Z factorial is one okay so here we have zero and a is one four this is uh uh this is La plus of e to X and this one here note that when we have n = 1 that means we have x e and a is one we have x e to a x this means that this is just L of x e x and using linearity we get e to x + x e to X and this means that Y is uh simply e to x + x e to X which is 1 + x e ^ x and this is the solution for our differential differential equation okay we can go back so one y is z so we have 0 a 0 which is one and if we integrate we're going to get this solution that we that we that we need we're going to get exactly this one other than that if we solve it the usual way we would get the same solution therefore using the plus transform we have arrived this is not minus this is uh so this is the solution that we get so 1 + xanti * e^ X and this is the solution of our differential equations and the key is learn how to use these General formulas when you have to like in our case and I think the best option is try to solve the pro this problem by yourself be active Okay be active use paper and pencil and Pen be active that's the key be active use the your handwriting and before to watch the video or maybe stop at every step and see if you can proceed and if you if you see any mistakes or errors please let us know and I would like to thank you you very much please share and subscribe thank you again

Channel: Archimedes Notes

Views: 9

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Id: KhtiT40AHp0

Length: 6min 4sec (364 seconds)

Published: Sat Jul 13 2024