- #1
blinktx411
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First time poster, so please feel free to leave any comments of a general nature.
I'm hoping to get a further insight on the derivation of the variation of parameters method used in ordinary differential equations to solve linear second order equations. I understand were looking for a particular solution of the form
[tex] y_{p}=v_{1}(t) \cdot y_1(t)+v_2(t) \cdot y_2(t) [/tex]
Taking the derivative with respect to t yields
[tex] \dfrac{d}{dt}y_p=(y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt}v_2) + (v_1 \cdot \dfrac{d}{dt}y_1 + v_2 \cdot \dfrac{d}{dt}y_2)[/tex]
The next step is to impose the requirement that
[tex] y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt} v_2 = 0[/tex]
It is at this point where I am not understanding (actually, this is the only step I don't understand in the derivation) . While I see that having the second derivatives of the functions [tex]v_1 [/itex] and [tex]v_2 [/itex] is going to be problematic, I do not understand why this requirement can be imposed without further explanation beyond "To simplify computation and to avoid second-order derivatives for the unknowns v1 and v2in the expression y''_p, we impose the requirement" (looked at a couple derivations online and in my old textbook, quote is from my textbook right before they impose the requirement I don't understand). Perhaps it is beyond the scope of the course, or perhaps I'm overlooking something blatantly simple? Any insight is appreciated.[/tex][/tex]
I'm hoping to get a further insight on the derivation of the variation of parameters method used in ordinary differential equations to solve linear second order equations. I understand were looking for a particular solution of the form
[tex] y_{p}=v_{1}(t) \cdot y_1(t)+v_2(t) \cdot y_2(t) [/tex]
Taking the derivative with respect to t yields
[tex] \dfrac{d}{dt}y_p=(y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt}v_2) + (v_1 \cdot \dfrac{d}{dt}y_1 + v_2 \cdot \dfrac{d}{dt}y_2)[/tex]
The next step is to impose the requirement that
[tex] y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt} v_2 = 0[/tex]
It is at this point where I am not understanding (actually, this is the only step I don't understand in the derivation) . While I see that having the second derivatives of the functions [tex]v_1 [/itex] and [tex]v_2 [/itex] is going to be problematic, I do not understand why this requirement can be imposed without further explanation beyond "To simplify computation and to avoid second-order derivatives for the unknowns v1 and v2in the expression y''_p, we impose the requirement" (looked at a couple derivations online and in my old textbook, quote is from my textbook right before they impose the requirement I don't understand). Perhaps it is beyond the scope of the course, or perhaps I'm overlooking something blatantly simple? Any insight is appreciated.[/tex][/tex]