Question Number 65601 by AnjanDey last updated on 31/Jul/19
![1.If y=x^(n−1) log x,then prove that,x^2 (d^2 y/dx^2 )+(3−2n)x(dy/dx)+(n−1)^2 y=0 2.If ((mtan (α−θ))/(cos^2 θ))=((ntan θ)/(cos^2 (α−θ))),then prove that,θ=(1/2)[α−tan^(−1) (((n−m)/(n+m))tan α)]](Q65601.png)
$$\mathrm{1}.\mathrm{If}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\boldsymbol{{n}}−\mathrm{1}} \mathrm{log}\:\boldsymbol{{x}},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\boldsymbol{{x}}^{\mathrm{2}} \frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }+\left(\mathrm{3}−\mathrm{2}\boldsymbol{{n}}\right)\boldsymbol{{x}}\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\left(\boldsymbol{{n}}−\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{{y}}=\mathrm{0} \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{I}}\mathrm{f}\:\frac{\mathrm{mtan}\:\left(\alpha−\theta\right)}{\mathrm{cos}\:^{\mathrm{2}} \theta}=\frac{{n}\mathrm{tan}\:\theta}{\mathrm{cos}\:^{\mathrm{2}} \left(\alpha−\theta\right)},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\theta=\frac{\mathrm{1}}{\mathrm{2}}\left[\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{n}−{m}}{{n}+{m}}\mathrm{tan}\:\alpha\right)\right] \\ $$
Commented by Prithwish sen last updated on 01/Aug/19
![2. ((sin(α−θ)cos(α−θ))/(sinθcosθ)) =(n/m)⇒((sin2(α−θ))/(sin2θ)) =(n/m) ⇒((sin2(α−θ)−sin2θ)/(sin2(α−θ)+sin2θ)) = ((n−m)/(n+m)) ⇒((tan(α−2θ))/(tanα)) =((n−m)/(n+m)) 1.(dy/dx) = (n−1)x^(n−2) logx + x^(n−2) (d^2 y/dx^2 ) =(n−1)(n−2)x^(n−3) logx+(n−1)x^(n−3) +(n−2)x^(n−3) x^2 (d^2 y/dx^2 ) = (n−1)(n−2)x^(n−1) logx + (n−1+n−2)x^(n−1) =− (n−1)^2 x^(n−1) logx+(n−1)(2n−3)x^(n−1) logx+(2n−3)x^(n−1) =−(n−1)^2 y−(3−2n)x[(n−1)x^(n−2) logx+x^(n−2) ] ⇒x^2 (d^2 y/dx^2 ) +(3−2n)(dy/dx) +(n−1)^2 y = 0 please check.](Q65647.png)
$$\mathrm{2}.\:\:\frac{\mathrm{sin}\left(\alpha−\theta\right)\mathrm{cos}\left(\alpha−\theta\right)}{\mathrm{sin}\theta\mathrm{cos}\theta}\:=\frac{\mathrm{n}}{\mathrm{m}}\Rightarrow\frac{\mathrm{sin2}\left(\alpha−\theta\right)}{\mathrm{sin2}\theta}\:=\frac{\mathrm{n}}{\mathrm{m}} \\ $$$$\Rightarrow\frac{\mathrm{sin2}\left(\alpha−\theta\right)−\mathrm{sin2}\theta}{\mathrm{sin2}\left(\alpha−\theta\right)+\mathrm{sin2}\theta}\:=\:\frac{\mathrm{n}−\mathrm{m}}{\mathrm{n}+\mathrm{m}} \\ $$$$\Rightarrow\frac{\mathrm{tan}\left(\alpha−\mathrm{2}\theta\right)}{\mathrm{tan}\alpha}\:=\frac{\mathrm{n}−\mathrm{m}}{\mathrm{n}+\mathrm{m}}\: \\ $$$$\mathrm{1}.\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}−\mathrm{2}} \mathrm{logx}\:+\:\mathrm{x}^{\mathrm{n}−\mathrm{2}} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)\mathrm{x}^{\mathrm{n}−\mathrm{3}} \mathrm{logx}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}−\mathrm{3}} +\left(\mathrm{n}−\mathrm{2}\right)\mathrm{x}^{\mathrm{n}−\mathrm{3}} \\ $$$$\mathrm{x}^{\mathrm{2}} \frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)\mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{logx}\:+\:\left(\mathrm{n}−\mathrm{1}+\mathrm{n}−\mathrm{2}\right)\mathrm{x}^{\mathrm{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\:\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{logx}+\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{2n}−\mathrm{3}\right)\mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{logx}+\left(\mathrm{2n}−\mathrm{3}\right)\mathrm{x}^{\mathrm{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{y}−\left(\mathrm{3}−\mathrm{2n}\right)\mathrm{x}\left[\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}−\mathrm{2}} \mathrm{logx}+\mathrm{x}^{\mathrm{n}−\mathrm{2}} \right] \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:+\left(\mathrm{3}−\mathrm{2n}\right)\frac{\mathrm{dy}}{\mathrm{dx}}\:+\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{y}\:=\:\mathrm{0} \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$