Applying the chain rule for a multivariate function
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?
differential-equations
asked 2 days ago
Neymar
374113
add a comment |
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?
differential-equations
asked 2 days ago
Neymar
374113
add a comment |
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?
differential-equations
asked 2 days ago
Neymar
374113
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?
differential-equations
differential-equations
asked 2 days ago
Neymar
374113
asked 2 days ago
Neymar
374113
asked 2 days ago
Neymar
374113
asked 2 days ago
Neymar
374113
asked 2 days ago
Neymar
374113
374113
add a comment |
add a comment |
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