问题及解答

求

\[
(x^2-1)dy+(2xy-\cos x)dx=0,
\]

满足 $y(0)=1$ 的特解.

记 $M(x,y)=2xy-\cos x$, $N(x,y)=x^2-1$. 则所考虑的方程满足

\[
\frac{\partial}{\partial y}M(x,y)=2x=\frac{\partial}{\partial x}N(x,y).
\]

于是存在函数 $u(x,y)$, 使得 $du(x,y)=M(x,y)dx+N(x,y)dy$.

根据公式

\[
u(x,y)=\int M(x,y)dx+\int\Bigl[N(x,y)-\frac{\partial}{\partial y}\int M(x,y)dx\Bigr]dy=C,
\]

得

\[
\begin{split}
u&=\int(2xy-\cos x)dx+\int\Bigl[(x^2-1)-\frac{\partial}{\partial y}\int(2xy-\cos x)dx\Bigr]dy\\
&=x^2 y-\sin x+\int\Bigl[(x^2-1)-\frac{\partial}{\partial}(x^2 y-\sin x)\Bigr]dy\\
&=x^2 y-\sin x-y,
\end{split}
\]

即 $u(x,y)=(x^2-1)y-\sin x=C$. 由条件 $y(0)=1$, 故得 $C=-1$. 所以解为

\[
y=\frac{\sin x-1}{x^2-1}.
\]