If y=acos⁡(log⁡x)+bsin⁡(log⁡x), where a & b are parameters, then x^2y^(")+xy^(')is equal to

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MARKS Content Team · Accepted Answer

Correct answer: (B) -y. y=acos⁡(log⁡x)+bsin⁡(log⁡x) On differentiating w.r.t. x, we get y^(′) =(−asin(logx))/x +(bcos(logx))/x ⇒xy^(′) =−asin(logx)+bcos(logx) Again, on differentiating w.r.t. x, we get xy^(″) +y^(′) =−acos(logx)1/x −bsin(logx)1/x ⇒x^2 y^(″) +y^(′) x=− acos(logx)+bsin(logx ] ⇒x^2 y^(″) +xy^(′) =−y

If $y = a \cos (\log x) + b \sin (\log x)$ , where $a & b$ are parameters, then $x^{2} y^{"} + x y^{'}$ is equal to

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More Differentiation Questions

If y=acos⁡(log⁡x)+bsin⁡(log⁡x), where a & b are parameters, then x2y"+xy'is equal to

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More Differentiation Questions

If $y = a \cos (\log x) + b \sin (\log x)$ , where $a & b$ are parameters, then $x^{2} y^{"} + x y^{'}$ is equal to