Differentiate it w.r.t. $x$, we get
$\frac{{dy}}{{dx}} = - {c_1}a\sin ax + {c_2}a\cos ax$
Again $\frac{{{d^2}y}}{{d{x^2}}} = - {c_1}{a^2}\cos ax - {c_2}{a^2}\sin ax$
$\frac{{{d^2}y}}{{d{x^2}}} = - {a^2}({c_1}\cos ax + {c_2}\sin ax) \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = - {a^2}y$
or $\frac{{{d^2}y}}{{d{x^2}}} + {a^2}y = 0$.
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$f\left( x \right)\left\{ \begin{array}{l}
\frac{{2{x^2}}}{a}\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,0 \le x < 1\,\,\,\\
a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,1 \le x < \sqrt 2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\frac{{2{b^2} - 4b}}{{{x^3}}}\,\,\,,\,\,\,\,\,\sqrt 2 \le x < \infty
\end{array} \right.\,\,\,\,$
is continuous in the interval $\left[ {0,\infty } \right)$ , then an ordered pair $(a, b)$ is