Applying the chain rule for a multivariate function












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$begingroup$


Suppose we have the system



$$ z_k' = f_k(x,z_1,...,z_n) $$



of first-order ODE. I want to differentiate this equation j times. IF we differentiate with respect to $x$ we have



$$ z_1'' = frac{ d }{dx} f_1(x,z_1,...,z_n) $$



by chain rule we have that RHS is



$$ frac{ partial f_1 }{partial x } x' + frac{ partial f_1 }{partial z_1} z_1' + ... + frac{ partial f_1 }{partial z_n } z_n' = f_1^{(1)}(x,z_1,...,z_n)$$



Now the jth derivative will also give a function of x and the $z's$ say $f^{(j-1)}_j (x,z_1,...,z_n)$



Now, my book says that if we consider the n equations $f_j^{j-1} = z_1^{(j)}$ then we can solve to obtain an equation of the form



$$ p_0 y + p_1 y'' + p_2 y''' + ... + y^{(n)} = f(x) $$



How is this possible?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Suppose we have the system



    $$ z_k' = f_k(x,z_1,...,z_n) $$



    of first-order ODE. I want to differentiate this equation j times. IF we differentiate with respect to $x$ we have



    $$ z_1'' = frac{ d }{dx} f_1(x,z_1,...,z_n) $$



    by chain rule we have that RHS is



    $$ frac{ partial f_1 }{partial x } x' + frac{ partial f_1 }{partial z_1} z_1' + ... + frac{ partial f_1 }{partial z_n } z_n' = f_1^{(1)}(x,z_1,...,z_n)$$



    Now the jth derivative will also give a function of x and the $z's$ say $f^{(j-1)}_j (x,z_1,...,z_n)$



    Now, my book says that if we consider the n equations $f_j^{j-1} = z_1^{(j)}$ then we can solve to obtain an equation of the form



    $$ p_0 y + p_1 y'' + p_2 y''' + ... + y^{(n)} = f(x) $$



    How is this possible?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose we have the system



      $$ z_k' = f_k(x,z_1,...,z_n) $$



      of first-order ODE. I want to differentiate this equation j times. IF we differentiate with respect to $x$ we have



      $$ z_1'' = frac{ d }{dx} f_1(x,z_1,...,z_n) $$



      by chain rule we have that RHS is



      $$ frac{ partial f_1 }{partial x } x' + frac{ partial f_1 }{partial z_1} z_1' + ... + frac{ partial f_1 }{partial z_n } z_n' = f_1^{(1)}(x,z_1,...,z_n)$$



      Now the jth derivative will also give a function of x and the $z's$ say $f^{(j-1)}_j (x,z_1,...,z_n)$



      Now, my book says that if we consider the n equations $f_j^{j-1} = z_1^{(j)}$ then we can solve to obtain an equation of the form



      $$ p_0 y + p_1 y'' + p_2 y''' + ... + y^{(n)} = f(x) $$



      How is this possible?










      share|cite|improve this question









      $endgroup$




      Suppose we have the system



      $$ z_k' = f_k(x,z_1,...,z_n) $$



      of first-order ODE. I want to differentiate this equation j times. IF we differentiate with respect to $x$ we have



      $$ z_1'' = frac{ d }{dx} f_1(x,z_1,...,z_n) $$



      by chain rule we have that RHS is



      $$ frac{ partial f_1 }{partial x } x' + frac{ partial f_1 }{partial z_1} z_1' + ... + frac{ partial f_1 }{partial z_n } z_n' = f_1^{(1)}(x,z_1,...,z_n)$$



      Now the jth derivative will also give a function of x and the $z's$ say $f^{(j-1)}_j (x,z_1,...,z_n)$



      Now, my book says that if we consider the n equations $f_j^{j-1} = z_1^{(j)}$ then we can solve to obtain an equation of the form



      $$ p_0 y + p_1 y'' + p_2 y''' + ... + y^{(n)} = f(x) $$



      How is this possible?







      ordinary-differential-equations






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 25 '18 at 18:14









      NeymarNeymar

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      406214






















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