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Let $$y''(x)+a(x)y'(x)+b(x)y(x)=f(x),$$ an ODE of second order. Let $y_1$ and $y_2$ two solution of the homogeneous equation. To find a general solution, we use variation constant method. Why do we have to find $y$ s.t. $$\begin{cases} y(x)=\lambda (x)y_1(x)+\mu(x)y_2(x)\\ y'(x)=\lambda (x)y_1'(x)+\mu(x)y_2'(x) \end{cases} ?$$

Shouldn't it be $$\begin{cases} y(x)=\lambda (x)y_1(x)+\mu(x)y_2(x)\\ y'(x)=\lambda (x)y_1'(x)+\lambda '(x)y_1(x)+\mu(x)y_2(x)+\mu'(x)y_2(x) \end{cases} ?$$ I mean, why is $y'(x)$ is not the derivative of $y(x)=\lambda (x)y_1(x)+\mu(x)y_2(x)$ ? Isn't it weird ?

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3 Answers 3

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Because we impose on $\lambda$ and $\mu$ the condition $$ \lambda'y_1+\mu'y_2=0. $$

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  • $\begingroup$ Thank you for your answer. But why do we impose such a condition ? I can see that, but I don't understand why... $\endgroup$
    – user330587
    Mar 10, 2018 at 12:10
  • $\begingroup$ You have to find two unknown functions: $\lambda$ and $\mu$. The idea is to obtain a system of two equations for them. The first equation is the one in the answer. Why that particular equation? Because it simplifies the calculations; in particular, it gives a simpler formula for y'. The second equation will come from $y''$. $\endgroup$ Mar 10, 2018 at 12:26
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(note that you miss the derivative on $y_2(x)$)

In the variation of constants method we start by assuming that: $y_1,y_2,\cdots,y_n$ is the fundamental system of solution for the homogeneous equation and that the particular solution is $\sum_{i=1}^nc_iy_i$ such that $$\sum_{i=1}^nc'y_i^{(j)}=0\text{ for all $j$ between $0$ and $n-2$}\quad(*)$$

So in your case $(*)$ is only for $j=0$ and $c_1,c_2$ are $\lambda,\mu$. So by assumption $\lambda '(x)y_1(x)+\mu'(x)y_2(x)=0$

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For the method of variation of parameters: If $y(x)=\lambda (x)y_1(x)+\mu(x)y_2(x)$,where $y_1$ and $y_2$ are fundamental solution of homogeneous part of the $2$nd order ODE, then must be $$y'(x)=\lambda (x)y_1'(x)+\lambda '(x)y_1(x)+\mu(x)y_2(x)+\mu'(x)y_2(x).$$ See wiki link where they say: Since $y(x)=\lambda (x)y_1(x)+\mu(x)y_2(x)$ is only one equation and we have two unknown functions, it is reasonable to impose a second condition ${\displaystyle \lambda'(x)y_{1}(x)+\lambda'(x)y_{2}(x)=0}$.

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