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(Page 125 of file, page 106, problem 19 of the book - Linear Analysis)
Hi folks,
I tried to come up with a proof for the following statement. But, I am facing trouble proceeding with the arguments. It would be nice, if someone could help. Pardon me for posting the question here, as there's no math sub-forum here.
Q. Show that two distinct solutions of a normal first order linear differential equation cannot have a point of intersection.
Proof.
Let \(L=a_{1}(x)D+a_{0}(x)\) be the linear differential operator of the first order and let
\[ Ly=h \]
be the given differential equation.
Suppose \(f(x)\) and \(g(x)\) are two distinct solutions of the linear differential equation, such that
\[Lf=h, \text{whenever }y(x_1)=k_1\]
\[Lg=h, \text{whenever }y(x_2)=k_2\]
But, I don't know where to go from here. Any tips on how to proceed would help me.
Thanks guys!
Hi folks,
I tried to come up with a proof for the following statement. But, I am facing trouble proceeding with the arguments. It would be nice, if someone could help. Pardon me for posting the question here, as there's no math sub-forum here.
Q. Show that two distinct solutions of a normal first order linear differential equation cannot have a point of intersection.
Proof.
Let \(L=a_{1}(x)D+a_{0}(x)\) be the linear differential operator of the first order and let
\[ Ly=h \]
be the given differential equation.
Suppose \(f(x)\) and \(g(x)\) are two distinct solutions of the linear differential equation, such that
\[Lf=h, \text{whenever }y(x_1)=k_1\]
\[Lg=h, \text{whenever }y(x_2)=k_2\]
But, I don't know where to go from here. Any tips on how to proceed would help me.
Thanks guys!
