Problem:
Consider the differential equation
$ y''' + y'y +y^2 = \cos(t) $
which is the correct conversion of this equation into first-order form
Consider also the boundary conditions
$ y′′(0)=5,y(0)=3,y(1)y′(1)+y(1)=0 $
What is the correct implementation of these boundary conditions in the first-order system.
Solution:
Let $ y_1 = y $, $ y_2 = y' $, $ y_3 = y'' $, we have
$ y_1' = y_2 $
$ y_2' = y_3 $
$ y_3' = y''' = \cos(t) - y'y - y^2 = \cos(t) - y_1y_2 - y_1^2 $
$ y''(0) = 5 \implies y_3(0) = 5 $
$ y(0) = 3 \implies y_1(0) = 3 $
$ y(1)y′(1)+y(1)=0 \implies y_1(1)y_2(1) +y_1(1) = 0 $
Consider the differential equation
$ y''' + y'y +y^2 = \cos(t) $
which is the correct conversion of this equation into first-order form
Consider also the boundary conditions
$ y′′(0)=5,y(0)=3,y(1)y′(1)+y(1)=0 $
What is the correct implementation of these boundary conditions in the first-order system.
Solution:
Let $ y_1 = y $, $ y_2 = y' $, $ y_3 = y'' $, we have
$ y_1' = y_2 $
$ y_2' = y_3 $
$ y_3' = y''' = \cos(t) - y'y - y^2 = \cos(t) - y_1y_2 - y_1^2 $
$ y''(0) = 5 \implies y_3(0) = 5 $
$ y(0) = 3 \implies y_1(0) = 3 $
$ y(1)y′(1)+y(1)=0 \implies y_1(1)y_2(1) +y_1(1) = 0 $
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