3、已知y_(1)(x)=e^x^(2)是微分方程y''+p(x)y'+q(x)y=0的一个解，设y_(2)(x)=v(x)y_(1)(x)是此方程另一个线性无关的解，则v(x)满足的微分方程是（）A. v''+(2x+p(x))v'=0B. v''+(4x+p(x))v'=0C. v''+(p(x)-2x)v'+q(x)v=0D. v''+(p(x)+4x)v'+q(x)v=0

本题考查二阶线性齐次微分方程解的结构以及通过已知解求另一个线性无关解所满足的方程，解题思路是先求出$y_{2}(x)=v(x)y_{1}(x)$的一阶导数和二阶导数，再将$y_{2}(x)$、$y_{2}'(x)$、$y_{2}''(x)$代入原微分方程，然后化简得到$v(x)$满足的微分方程。

步骤一：求$y_{2}(x)$的一阶导数$y_{2}'(x)$

已知$y_{2}(x)=v(x)y_{1}(x)$，且$y_{1}(x)=e^{x^{2}}$，根据乘积的求导法则$(uv)^\prime = u^\prime v + uv^\prime$，可得：
$y_{2}'(x)=v'(x)y_{1}(x)+v(x)y_{1}'(x)$
对$y_{1}(x)=e^{x^{2}}$求导，根据复合函数求导法则$(e^{u})^\prime = e^{u} \cdot u^\prime$，令$u = x^2$，则$y_{1}'(x)=e^{x^{2}}\cdot 2x$，所以：
$y_{2}'(x)=v'(x)e^{x^{2}}+2xv(x)e^{x^{2}}=e^{x^{2}}(v'(x)+2xv(x))$

步骤二：求$y_{2}(x)$的二阶导数$y_{2}''(x)$

对$y_{2}'(x)=e^{x^{2}}(v'(x)+2xv(x))$再次使用乘积的求导法则求导：
$y_{2}''(x)=(e^{x^{2}})^\prime(v'(x)+2xv(x))+e^{x^{2}}(v'(x)+2xv(x))^\prime$
由前面求导可知$(e^{x^{2}})^\prime = 2xe^{x^{2}}$，对$(v'(x)+2xv(x))$求导得$(v'(x)+2xv(x))^\prime = v''(x)+2v(x)+2xv'(x)$，所以：
$\begin{align*}y_{2}''(x)&=2xe^{x^{2}}(v'(x)+2xv(x))+e^{x^{2}}(v''(x)+2v(x)+2xv'(x))\\&=e^{x^{2}}(v''(x)+4xv'(x)+4x^{2}v(x)+2v(x))\end{align*}$

步骤三：将$y_{2}(x)$、$y_{2}'(x)$、$y_{2}''(x)$代入原微分方程

将$y_{2}(x)=v(x)e^{x^{2}}$、$y_{2}'(x)=e^{x^{2}}(v'(x)+2xv(x))$、$y_{2}''(x)=e^{x^{2}}(v''(x)+4xv'(x)+4x^{2}v(x)+2v(x))$代入原方程$y''+p(x)y'+q(x)y=0$得：
$\begin{align*}e^{x^{2}}(v''(x)+4xv'(x)+4x^{2}v(x)+2v(x))+p(x)e^{x^{2}}(v'(x)+2xv(x))+q(x)v(x)e^{x^{2}}&=0\end{align*}$
因为$e^{x^{2}}\neq 0$，两边同时除以$e^{x^{2}}$得：
$v''(x)+4xv'(x)+4x^{2}v(x)+2v(x)+p(x)v'(x)+2xp(x)v(x)+q(x)v(x)=0$
整理得：
$v''(x)+(4x + p(x))v'(x)+(4x^{2}+2 + 2xp(x)+q(x))v(x)=0$
由于$y_{1}(x)=e^{x^{2}}$是原方程的解，将$y_{1}(x)=e^{x^{2}}$代入原方程可得：
$(e^{x^{2}})''+p(x)(e^{x^{2}})'+q(x)e^{x^{2}}=0$
即$(4x^{2}+2)e^{x^{2}}+2xp(x)e^{x^{2}}+q(x)e^{x^{2}}=0$，两边同时除以$e^{x^{2}}$得$4x^{2}+2 + 2xp(x)+q(x)=0$。
将$4x^{2}+2 + 2xp(x)+q(x)=0$代入$v''(x)+(4x + p(x))v'(x)+(4x^{2}+2 + 2xp(x)+q(x))v(x)=0$，可得：
$v''(x)+(4x + p(x))v'(x)=0$