Find the fundamental set of solutions for $-y''+k^2y=f$ with the boundary conditions $2y(0)-y'(0)=0$ and...
I am reading a chapter from a textbook about Green's functions, and the example in the text asks to find the Green's function for $-y''+k^2y=f$ with the boundary conditions $2y(0)-y'(0)=0$ and $y(1)=0$.
The solution starts by saying to take $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$ and then concludes by giving the corresponding Green's function.
I can't seem to figure out how the author was able to obtain those fundamental solutions.
I know that the general solution to the homogeneous case is given by $y(x) = c_1e^{kx} + c_2e^{-kx}$, but I don't understand how solving for $c_1$ and $c_2$ with the boundary conditions produces a $y_1(x)$ with two terms.
So my question is: how did the author go from a general solution to the homogeneous case being $y(x) = c_1e^{kx} + c_2e^{-kx}$ to obtaining the fundamental solutions $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$?
differential-equations greens-function
asked Dec 10 '18 at 3:07
playitright
337112
add a comment |
I am reading a chapter from a textbook about Green's functions, and the example in the text asks to find the Green's function for $-y''+k^2y=f$ with the boundary conditions $2y(0)-y'(0)=0$ and $y(1)=0$.
The solution starts by saying to take $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$ and then concludes by giving the corresponding Green's function.
I can't seem to figure out how the author was able to obtain those fundamental solutions.
I know that the general solution to the homogeneous case is given by $y(x) = c_1e^{kx} + c_2e^{-kx}$, but I don't understand how solving for $c_1$ and $c_2$ with the boundary conditions produces a $y_1(x)$ with two terms.
So my question is: how did the author go from a general solution to the homogeneous case being $y(x) = c_1e^{kx} + c_2e^{-kx}$ to obtaining the fundamental solutions $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$?
differential-equations greens-function
asked Dec 10 '18 at 3:07
playitright
337112
add a comment |
I am reading a chapter from a textbook about Green's functions, and the example in the text asks to find the Green's function for $-y''+k^2y=f$ with the boundary conditions $2y(0)-y'(0)=0$ and $y(1)=0$.
The solution starts by saying to take $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$ and then concludes by giving the corresponding Green's function.
I can't seem to figure out how the author was able to obtain those fundamental solutions.
I know that the general solution to the homogeneous case is given by $y(x) = c_1e^{kx} + c_2e^{-kx}$, but I don't understand how solving for $c_1$ and $c_2$ with the boundary conditions produces a $y_1(x)$ with two terms.
So my question is: how did the author go from a general solution to the homogeneous case being $y(x) = c_1e^{kx} + c_2e^{-kx}$ to obtaining the fundamental solutions $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$?
differential-equations greens-function
asked Dec 10 '18 at 3:07
playitright
337112
I am reading a chapter from a textbook about Green's functions, and the example in the text asks to find the Green's function for $-y''+k^2y=f$ with the boundary conditions $2y(0)-y'(0)=0$ and $y(1)=0$.
The solution starts by saying to take $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$ and then concludes by giving the corresponding Green's function.
I can't seem to figure out how the author was able to obtain those fundamental solutions.
I know that the general solution to the homogeneous case is given by $y(x) = c_1e^{kx} + c_2e^{-kx}$, but I don't understand how solving for $c_1$ and $c_2$ with the boundary conditions produces a $y_1(x)$ with two terms.
So my question is: how did the author go from a general solution to the homogeneous case being $y(x) = c_1e^{kx} + c_2e^{-kx}$ to obtaining the fundamental solutions $y_1(x) = 2sinh(kx)+kcosh(kx)$ and $y_2(x) = sinh(k(1-x))$?
differential-equations greens-function
differential-equations greens-function
asked Dec 10 '18 at 3:07
playitright
337112
asked Dec 10 '18 at 3:07
playitright
337112
asked Dec 10 '18 at 3:07
playitright
337112
asked Dec 10 '18 at 3:07
playitright
337112
asked Dec 10 '18 at 3:07
playitright
337112
337112
add a comment |
add a comment |
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