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Q.
If $y=x^5(\cos (\ln x)+\sin (\ln x))$, then find the value of $(a+b)$ in the relation $x^2 y_2+a x y_1+b y$ $=0$.
Continuity and Differentiability
Solution:
We have $\frac{ dy }{ dx }=5 x ^4(\cos (\ln x )+\sin (\ln x ))+ x ^5\left(\frac{-\sin (\ln x )}{ x }+\frac{\cos (\ln x )}{ x }\right)$,
$\Rightarrow x_1=5 y+x^5(\cos (\ln x)-\sin (\ln x)) $
$\Rightarrow xy _2+ y _1=5 y _1+5 x ^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=5 x y_1+5 x^5(\cos (\ln x)-\sin (\ln x))-x^5(\sin (\ln x)+\cos (\ln x)) $
$\Rightarrow x^2 y_2-4 x_1=5\left(x_1-5 y\right)-y$
$\Rightarrow x ^2 y _2-4 xy _1=5 xy _1-26 y$
$\Rightarrow x^2 y_2-9 x y_1+26 y=0 \equiv x^2 y_2+a x y_1+b y=0 $
$\therefore a=-9 \text { and } b=26$
$\text { Hence }(a+b)=17$