Question

If $y=x^5\left(\cos \right(\ln x)+\sin (\ln x\left)\right)$, then the value of $(a+b)$ in the relation $x^2{y}_2+ax{y}_1+by=0$ is
$(y_1$ and $y_2$ denote the first and second derivative of $y$ with respect to $x$, respectively.$)$

Answer 1

We have $y=x^5\left(\cos \right(\ln x)+\sin (\ln x\left)\right)$
$\frac{dy}{dx}=5x^4\left(\cos \right(\ln x)+\sin (\ln x\left)\right)+x^5(\frac{-\sin \left(\ln x\right)}{x}+\frac{\cos \left(\ln x\right)}{x})$
$\Rightarrow x{y}_1=5y+x^5\left(\cos \right(\ln x)-\sin (\ln x\left)\right)$
$\Rightarrow x{y}_2+y_1=5y_1+5x^4\left(\cos \right(\ln x)-\sin (\ln x\left)\right)+x^5(\frac{-\sin \left(\ln x\right)}{x}-\frac{\cos \left(\ln x\right)}{x})$
$\Rightarrow x^2{y}_2+x{y}_1=5x{y}_1+5x^5\left(\cos \right(\ln x)-\sin (\ln x\left)\right)-x^5\left(\sin \right(\ln x)+\cos (\ln x\left)\right)$
$\Rightarrow x^2{y}_1-4x{y}_1=5(x{y}_1-5y)-y$
$\Rightarrow x^2{y}_2-4x{y}_1=5x{y}_1-26y$
$\Rightarrow x^2{y}_2-9x{y}_1+26y=0$
Comparing with $x^2{y}_2+ax{y}_1+by=0,$ we get
$a=-9$ and $b=26$
Hence, $a+b=17$