Can you solve an ODE of type y′′′+y′′cdots=0 with Laplace transforms if you don't get the values of y_0?

restgarnut

restgarnut

Answered question

2022-09-10

Can you solve an ODE of type y + y = 0 with Laplace transforms if you don't get the values of y 0 ?

Answer & Explanation

Isaiah Haynes

Isaiah Haynes

Beginner2022-09-11Added 16 answers

Assume that y : [ 0 , ) R is n + 1-times differentiable. We compute
L ( y ( t ) ) = 0 e s t y ( t )   d t = e s t y ( t ) | 0 + s 0 e s t y ( t )   d t = Y ( s ) y ( 0 ) .
Assume now that L ( y ( n ) ( t ) ) = s n Y ( s ) j = 1 n y ( n j ) ( 0 ) s j 1 for some nonnegative integer n. Then
L ( y ( n + 1 ) ( t ) ) = 0 e s t y ( n + 1 ) ( t )   d t = e s t y ( n ) | 0 + s 0 e s t y ( n ) ( t )   d t = y ( n ) ( 0 ) + s L ( y ( n ) ( t ) ) = y ( n ) ( 0 ) + s ( s n Y ( s ) j = 1 n y ( n j ) ( 0 ) s j 1 = s n + 1 Y ( s ) + y ( n ) ( 0 ) j = 1 n y ( n j ) ( 0 ) s j = S n + 1 Y ( s ) + j = 1 n + 1 y n + 1 j ( 0 ) s j 1 ,
so by mathematical induction, the formula holds true for all nonnegative integers. Suppose i = 0 m a i y ( i ) ( t ) = 0, then
L ( i = 0 m a i y ( i ) ( t ) ) = i = 0 m a i L ( y ( i ) ( t ) = i = 0 m a i ( s i Y ( s ) j = 1 i y ( i j ) ( 0 ) s j 1 ) = Y ( s ) i = 0 m a i s i i = 0 m j = i y ( i j ) ( 0 ) s j 1 = Y ( s ) i = 0 m a i s i j = 1 m i = j m y ( i j ) ( 0 ) s j 1 = Y ( s ) i = 0 m a i s i j = 1 m s j 1 i = 0 m j y ( i ) ( 0 ) .
Now, since L ( i = 0 m a i y ( i ) ( t ) ) = 0, we have
Y ( s ) = j = 1 m s j 1 i = 0 m j y ( i ) ( 0 ) i = 0 m a i s i .
As you can see, we can compute the Laplace transform, but cannot solve the ODE without knowledge of the values of y ( i ) ( 0 )

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