- #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
<br /> y_{p}=v_{1}(t) \cdot y_1(t)+v_2(t) \cdot y_2(t) <br />
Taking the derivative with respect to t yields
<br /> \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)<br />
The next step is to impose the requirement that
<br /> y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt} v_2 = 0<br />
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 v_1 [/itex] and v_2 [/itex] is going to be problematic, I do not understand why this requirement can be imposed without further explanation beyond &quot;To simplify computation and to avoid second-order derivatives for the unknowns v1 and v2in the expression y&#039;&#039;_p, we impose the requirement&quot; (looked at a couple derivations online and in my old textbook, quote is from my textbook right before they impose the requirement I don&#039;t understand). Perhaps it is beyond the scope of the course, or perhaps I&#039;m overlooking something blatantly simple? Any insight is appreciated.
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
<br /> y_{p}=v_{1}(t) \cdot y_1(t)+v_2(t) \cdot y_2(t) <br />
Taking the derivative with respect to t yields
<br /> \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)<br />
The next step is to impose the requirement that
<br /> y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt} v_2 = 0<br />
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 v_1 [/itex] and v_2 [/itex] is going to be problematic, I do not understand why this requirement can be imposed without further explanation beyond &quot;To simplify computation and to avoid second-order derivatives for the unknowns v1 and v2in the expression y&#039;&#039;_p, we impose the requirement&quot; (looked at a couple derivations online and in my old textbook, quote is from my textbook right before they impose the requirement I don&#039;t understand). Perhaps it is beyond the scope of the course, or perhaps I&#039;m overlooking something blatantly simple? Any insight is appreciated.
