<transcript><audio_19>using integral which is a linear operation so you can split in this part equals zero which that's for the right-hand side you had already found a person's form of U double because square times u minus Omega and we add I'm gonna scratch</transcript><transcript><audio_20>U of S equals zero and then we can solve for u of s it's now an algebraic equation if a skirt was I'm gonna scratch times u of s equals Omega so we can solve for capital u of s so now it's</transcript><transcript><audio_21>that of solving differential equation we had to solve an algebraic question which was trivial in this case so we find the Laplace transform of the solution but how do we find our original function back well in this case we're lucky we</transcript><transcript><audio_22>found capital university cos Omega divided by s square plus Omega squared and we know the original function because we in our first example we found the Laplace transform of sine Omega T at exactly this function Omega divided by</transcript><transcript><audio_23>escarzaga skirt so isn't that lucky so now we can find our U of T immediately because it has to be sine of Omega T because a Laplace transform of sine Omega T was it exactly this over there</transcript><transcript><audio_24>so that's how you use the episode forms of soft differential equations let's take a look at the third example let's try to find a Laplace transform of other confusion we encounter very often functions of T to the power and and some</transcript><transcript><audio_25>positive power we start with the person for T to the power 0 which is basically 1 so we integrate its power I'm ste a star on the s trivial so I get a minus 1 over s to the power minus TS plug in the boundaries of course you</transcript><transcript><audio_26>need again the real part of has to be bigger than 0 in order to get rid of the upper boundary and you get 1 of s so that's 40 to power 0 what about some general power T to the power and bother you have T to the power and over here</transcript> using integral which is a linear operation so you can split in this part equals zero which that's for the right-hand side you had already found a person's form of U double because square times u minus Omega and we add I'm gonna scratch U of S equals zero and then we can solve for u of s it's now an algebraic equation if a skirt was I'm gonna scratch times u of s equals Omega so we can solve for capital u of s so now it's that of solving differential equation we had to solve an algebraic question which was trivial in this case so we find the Laplace transform of the solution but how do we find our original function back well in this case we're lucky we found capital university cos Omega divided by s square plus Omega squared and we know the original function because we in our first example we found the Laplace transform of sine Omega T at exactly this function Omega divided by escarzaga skirt so isn't that lucky so now we can find our U of T immediately because it has to be sine of Omega T because a Laplace transform of sine Omega T was it exactly this over there so that's how you use the episode forms of soft differential equations let's take a look at the third example let's try to find a Laplace transform of other confusion we encounter very often functions of T to the power and and some positive power we start with the person for T to the power 0 which is basically 1 so we integrate its power I'm ste a star on the s trivial so I get a minus 1 over s to the power minus TS plug in the boundaries of course you need again the real part of has to be bigger than 0 in order to get rid of the upper boundary and you get 1 of s so that's 40 to power 0 what about some general power T to the power and bother you have T to the power and over here
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