5. Given real numbers $a, b, c$, and $b \neq 0$. If real numbers $x_{1},$ $x_{2}, y_{1}, y_{2}$ satisfy $x_{1}^{2}+a x_{2}^{2}=b, x_{2} y_{1}-x_{1} y_{2}=a$, $x_{1} y_{1}+a x_{2} y_{2}=c$, then the value of $y_{1}^{2}+a y_{2}^{2}$ is $\qquad$
5. $\frac{c^{2}+a^{3}}{b}$.

From the given, we have $\left(x_{2} y_{1}-x_{1} y_{2}\right)^{2}=a^{2}$, that is
$$
\begin{array}{l}
2 x_{1} x_{2} y_{1} y_{2}+a^{2}=x_{2}^{2} y_{1}^{2}+x_{1}^{2} y_{2}^{2} . \\
\text { Hence } b\left(y_{1}^{2}+a y_{2}^{2}\right)=\left(x_{1}^{2}+a x_{2}^{2}\right)\left(y_{1}^{2}+a y_{2}^{2}\right) \\
=x_{1}^{2} y_{1}^{2}+a\left(x_{1}^{2} y_{2}^{2}+x_{2}^{2} y_{1}^{2}\right)+a^{2} x_{2}^{2} y_{2}^{2} \\
=x_{1}^{2} y_{1}^{2}+a\left(2 x_{1} x_{2} y_{1} y_{2}+a^{2}\right)+a^{2} x_{2}^{2} y_{2}^{2} \\
=\left(x_{1} y_{1}+a x_{2} y_{2}\right)^{2}+a^{3}=c^{2}+a^{3} .
\end{array}
$$

Since $b \neq 0$, we have $y_{1}^{2}+a y_{2}^{2}=\frac{c^{2}+a^{3}}{b}$.