How does the Wronskian determine uniqueness of solution?

[MathewsMD]

Given an nth order DE, how (intuitively and/or mathematically) does computing the Wronskian to be nonzero for at least one point in the defined interval for the solution to the DE ensure the solution is unique and also a fundamental set of solutions?

Also, is it true that if W = 0, it is 0 for all entries, and also if W does not equal 0, it is never equal to 0? If so, is there a formal proof (and/or intuitive explanation) for this?

Also, for a non-homogeneous DE, the solution is in the form Yhomog + Ypart = y, where Ypart is a solution to the g(t). Does Ypart necessarily have to have ALL solutions or just one?

 

[LCKurtz]

Given an nth order DE, how (intuitively and/or mathematically) does computing the Wronskian to be nonzero for at least one point in the defined interval for the solution to the DE ensure the solution is unique and also a fundamental set of solutions?

Also, is it true that if W = 0, it is 0 for all entries, and also if W does not equal 0, it is never equal to 0? If so, is there a formal proof (and/or intuitive explanation) for this?

You want to look at Abel's identity. Look here: http://archive.lib.msu.edu/crcmath/math/math/a/a009.htm
Basically it shows the Wronskian is a constant times an exponential type function, so it's always zero or never zero.

Also, for a non-homogeneous DE, the solution is in the form Yhomog + Ypart = y, where Ypart is a solution to the g(t). Does Ypart necessarily have to have ALL solutions or just one?

Call your DE $L(y) = f(x)$. If it is a second order DE then $y_c = Cy_1 + Dy_2$ is the general solution of the homogeneous equation. Suppose $y_p$ is a particular solution of the NH equation $L(y_p) =f(x)$. Then if $y_q$ is any solution of the NH equation you have $L(y_q)=f(x)$ so $L(y_q-y_p) = f(x) - f(x) = 0$. This says $y_q-y_p$ satisfies the homogeneous equation so there are values of $C$ and $D$ such that $y_q-y_p=Cy_1+Dy_2$. This tells you that $y_q = y_p + Cy_1+Dy_2$. Since $y_q$ could be any solution of the DE, this shows that your original $y_p,~ y_1,~y_2$ are good enough to express any solution.

 

[MathewsMD]

You want to look at Abel's identity. Look here: http://archive.lib.msu.edu/crcmath/math/math/a/a009.htm
Basically it shows the Wronskian is a constant times an exponential type function, so it's always zero or never zero.
Call your DE $L(y) = f(x)$. If it is a second order DE then $y_c = Cy_1 + Dy_2$ is the general solution of the homogeneous equation. Suppose $y_p$ is a particular solution of the NH equation $L(y_p) =f(x)$. Then if $y_q$ is any solution of the NH equation you have $L(y_q)=f(x)$ so $L(y_q-y_p) = f(x) - f(x) = 0$. This says $y_q-y_p$ satisfies the homogeneous equation so there are values of $C$ and $D$ such that $y_q-y_p=Cy_1+Dy_2$. This tells you that $y_q = y_p + Cy_1+Dy_2$. Since $y_q$ could be any solution of the DE, this shows that your original $y_p,~ y_1,~y_2$ are good enough to express any solution.

Thank you so much! That's exactly what I was looking for!