Question

If $y=x^5(\cos (\ln x)+\sin (\ln x))$, then find the value of $(a+b)$ in the relation $x^2{y}_2+ax{y}_1+by$ $=0$.

Answer 1

Answer: 17

We have $\frac{dy}{dx}=5x^4(\cos (\ln x)+\sin (\ln x))+x^5\left(\frac{-\sin (\ln x)}{x}+\frac{\cos (\ln x)}{x}\right)$,
$\Rightarrow x_1=5y+x^5(\cos (\ln x)-\sin (\ln x))$
$\Rightarrow x{y}_2+y_1=5y_1+5x^4(\cos (\ln x)-\sin (\ln x))+x^5\left(\frac{-\sin (\ln x)}{x}-\frac{\cos (\ln x)}{x}\right)$
$\Rightarrow x^2{y}_2+x{y}_1=5x{y}_1+5x^5(\cos (\ln x)-\sin (\ln x))-x^5(\sin (\ln x)+\cos (\ln x))$
$\Rightarrow x^2{y}_2-4x_1=5\left(x_1-5y\right)-y$
$\Rightarrow x^2{y}_2-4x{y}_1=5x{y}_1-26y$
$\Rightarrow x^2{y}_2-9x{y}_1+26y=0\equiv x^2{y}_2+ax{y}_1+by=0$
$\therefore a=-9\text{ and }b=26$
$\text{ Hence }(a+b)=17$