I am trying to solve a first order differential equation with the condition that g ( y ) = 0 if y = 0: a g ′ ( c y ) + b g ′ ( e y ) = α (1) g ( 0 ) = 0 , where parameters a,b,c,e are real nonzero constants; α is a complex constant; function g ( y ) : R → C is a function mapping from real number y to a complex number. The goal is to solve for function g ( ⋅ ). This is what I have done. Solve this differential equation by integrating with respect to y: a c g ( c y ) + b e g ( e y ) = α y + β , where β is another complex constant. Plugging in y=0 and using the fact that g(0)=0, we have β = 0. Therefore, we have a c g ( c y ) + b e g ( e y ) = α y . The background of this problem is Cauchy functional equation, so my conjecture is one solution could be g ( y ) = γ y. Plugging in g ( y ) = γ y, I get γ = α a + b , which implies that one solution is g ( y ) = α a + b y. Then, I move on to show uniqueness. I define a vector-valued function h ≡ ( h 1 , h 2 ) T such that h 1 ( y ) = a c g 1 ( c y ) + b e g 1 ( e y ) h 2 ( y ) = a c g 2 ( c y ) + b e g 2 ( e y ) , where g ( y ) ≡ g 1 ( y ) + i g 2 ( y ). Then, I rewrite this differential equation as h ′ ( y ) = α (2) h ( 0 ) = 0 , where α ≡ ( α 1 , i α 2 ) T . By the uniqueness theorem of first order differential equation, solution h(y) is unique. I have two questions. First, I think equation (1) and (2) should be equivalent. However, it seems that equation (1) can imply equation (2) but equation (2) may not imply equation (1). This is because h(0)=0 may imply either g 1 ( 0 ) = 0 , g 2 ( 0 ) = 0 or a c + b e = 0. Second, I have only proved that h(y) is unique. How should I proceed to show g(y) is also unique.

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I am trying to solve a first order differential equation with the condition that   g  (  y  )  =  0 if   y  =  0:                          a                  g          ′                (        c        y        )        +        b                  g          ′                (        e        y        )        =        α                            (1)                          g        (        0        )        =        0        ,            where parameters a,b,c,e are real nonzero constants;   α is a complex constant; function   g  (  y  )  :      R    →      C   is a function mapping from real number y to a complex number. The goal is to solve for function   g  (  ⋅  ). This is what I have done. Solve this differential equation by integrating with respect to y:                              a          c                g        (        c        y        )        +                  b          e                g        (        e        y        )        =        α        y        +        β        ,            where   β is another complex constant. Plugging in y=0 and using the fact that g(0)=0, we have   β  =  0. Therefore, we have                              a          c                g        (        c        y        )        +                  b          e                g        (        e        y        )        =        α        y        .            The background of this problem is Cauchy functional equation, so my conjecture is one solution could be   g  (  y  )  =  γ  y. Plugging in   g  (  y  )  =  γ  y, I get   γ  =      α          a      +      b      , which implies that one solution is   g  (  y  )  =      α          a      +      b        y. Then, I move on to show uniqueness. I define a vector-valued function   h  ≡  (      h    1    ,      h    2        )    T   such that                              h          1                (        y        )                            =                  a          c                          g          1                (        c        y        )        +                  b          e                          g          1                (        e        y        )                                      h          2                (        y        )                            =                  a          c                          g          2                (        c        y        )        +                  b          e                          g          2                (        e        y        )        ,            where   g  (  y  )  ≡      g    1    (  y  )  +  i      g    2    (  y  ). Then, I rewrite this differential equation as                                    h          ′                (        y        )        =        α                            (2)                          h        (        0        )        =        0        ,            where   α  ≡  (      α    1    ,  i      α    2        )    T  . By the uniqueness theorem of first order differential equation, solution h(y) is unique. I have two questions. First, I think equation (1) and (2) should be equivalent. However, it seems that equation (1) can imply equation (2) but equation (2) may not imply equation (1). This is because h(0)=0 may imply either       g    1    (  0  )  =  0  ,      g    2    (  0  )  =  0 or       a    c    +      b    e    =  0. Second, I have only proved that h(y) is unique. How should I proceed to show g(y) is also unique.

jugf5 · Accepted Answer

For people who are interested in my question, I got an answer from a friend. This is a smart way to show uniqueness of solution.                    a                  g          ′                (        c        y        )        +        b                  g          ′                (        e        y        )        =        α            Define   f  (  ⋅  )  ≡      g    ′    (  ⋅  ):                    a        f        (        c        y        )        +        b        f        (        e        y        )        =        α            View this equation as an infinite system of algebraic equations in unknowns (a,b). For example,                    a        f        (        0        )        +        b        f        (        0        )                            =        α                            a        f        (        c        )        +        b        f        (        e        )                            =        α                            a        f        (        2        c        )        +        b        f        (        2        e        )                            =        α                                          …            Since a and b are non-zero, this system of equations is solvable if and only if function   f  (  ⋅  ) is a constant function. Thus, function       g    ′    (  ⋅  ) is a constant function which further implies that function   g  (  y  )  =  β  y where   β is a complex constant.

I am trying to solve a first order differential equation with the condition that g ( y

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