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Question Number 83849 by Rio Michael last updated on 06/Mar/20

Gven that y = e^(−x) sinbx ,where b is a constant,show that (d^2 y/dx^2 ) + 2(dy/dx) + (1 + b^2 )y = 0.

$$\mathrm{Gven}\:\mathrm{that}\:{y}\:=\:{e}^{−{x}} \mathrm{sin}{bx}\:,\mathrm{where}\:{b}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant},\mathrm{show}\:\mathrm{that} \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\left(\mathrm{1}\:+\:{b}^{\mathrm{2}} \right){y}\:=\:\mathrm{0}. \\ $$

Commented by niroj last updated on 06/Mar/20

Given, y= e^(−x) sin bx....(i) (dy/dx)= −1.e^(−x) .sinbx+e^(−x) .b.cos bx (dy/dx)= −y+e^(−x) .bcosbx .....(ii) (∵ y=e^(−x) sin bx) (d^2 y/dx^2 )= −(dy/dx)+(−1.e^(−x) b cos bx−e^(−x) b^2 sinbx) (d^2 y/dx^2 )= −(dy/dx)−e^x bcosbx−b^2 y (d^2 y/dx^2 )= −(dy/dx)−((dy/dx)+y)−b^2 y { from..(ii)} (d^2 y/dx^2 )= −(dy/dx)−(dy/dx)−y−b^2 y (d^2 y/dx^2 )= −2(dy/dx)−(1+b^2 )y ∴ (d^2 y/dx^2 )+2(dy/dx)+(1+b^2 )y=0 hence proved//.

$$\:\: \\ $$$$\:\:\mathrm{Given}, \\ $$$$\:\:\:\:\mathrm{y}=\:\mathrm{e}^{−\mathrm{x}} \mathrm{sin}\:\mathrm{bx}....\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=\:−\mathrm{1}.\mathrm{e}^{−\mathrm{x}} .\mathrm{sinbx}+\mathrm{e}^{−\mathrm{x}} .\mathrm{b}.\mathrm{cos}\:\mathrm{bx} \\ $$$$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=\:−\mathrm{y}+\mathrm{e}^{−\mathrm{x}} .\mathrm{bcosbx}\:.....\left(\mathrm{ii}\right)\:\left(\because\:\mathrm{y}=\mathrm{e}^{−\mathrm{x}} \mathrm{sin}\:\mathrm{bx}\right) \\ $$$$\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\:−\frac{\mathrm{dy}}{\mathrm{dx}}+\left(−\mathrm{1}.\mathrm{e}^{−\mathrm{x}} \mathrm{b}\:\mathrm{cos}\:\mathrm{bx}−\mathrm{e}^{−\mathrm{x}} \mathrm{b}^{\mathrm{2}} \mathrm{sinbx}\right) \\ $$$$\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\:−\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{e}^{\mathrm{x}} \mathrm{bcosbx}−\mathrm{b}^{\mathrm{2}} \mathrm{y} \\ $$$$\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\:−\frac{\mathrm{dy}}{\mathrm{dx}}−\left(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right)−\mathrm{b}^{\mathrm{2}} \mathrm{y}\:\:\:\left\{\:\:\mathrm{from}..\left(\mathrm{ii}\right)\right\} \\ $$$$\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\:−\frac{\mathrm{dy}}{\mathrm{dx}}−\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{y}−\mathrm{b}^{\mathrm{2}} \mathrm{y} \\ $$$$\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\:−\mathrm{2}\frac{\mathrm{dy}}{\mathrm{dx}}−\left(\mathrm{1}+\mathrm{b}^{\mathrm{2}} \right)\mathrm{y} \\ $$$$\:\therefore\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{2}\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\mathrm{1}+\mathrm{b}^{\mathrm{2}} \right)\mathrm{y}=\mathrm{0}\:\:\mathrm{hence}\:\mathrm{proved}//. \\ $$

Commented by Rio Michael last updated on 06/Mar/20

thanks sir

$${thanks}\:{sir} \\ $$

Commented by niroj last updated on 07/Mar/20

you must welcome sir.

$${you}\:{must}\:{welcome}\:{sir}. \\ $$

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