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Question Number 86640 by niroj last updated on 29/Mar/20
Solve the differential equations: (i).x^2 (d^2 y/dx^2 ) − x(dy/dx) + y = log x. (ii). (x+2)^2 (d^2 y/dx^2 ) − 4(x+2)(dy/dx) + 6y = x.
$$\:\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equations}}: \\ $$$$\:\:\left(\boldsymbol{\mathrm{i}}\right).\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\:\boldsymbol{\mathrm{x}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:+\:\boldsymbol{\mathrm{y}}\:=\:\:\boldsymbol{\mathrm{log}}\:\boldsymbol{\mathrm{x}}. \\ $$$$\:\:\left(\boldsymbol{\mathrm{ii}}\right).\:\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)^{\mathrm{2}} \:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\:\mathrm{4}\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:+\:\mathrm{6}\boldsymbol{\mathrm{y}}\:=\:\:\boldsymbol{\mathrm{x}}. \\ $$$$\: \\ $$
Answered by mind is power last updated on 30/Mar/20
(i) homgenius x^2 y′′−xy′+y=0 y=x^t ⇒(t(t−1)−t+1))x^t =0 ⇒(t^2 −2t+1)x^t =0⇒t=1⇒y=x let y=xz⇒y′=z+xz′⇒y′′=2z′+xz′′ ⇒x^2 (xz′′+2z′)−x(z+xz′)+xz=0 ⇒x^2 (xz′′+z′)=0⇒xz′′+z′ ⇒ln(z′)=ln((1/x))+c⇒z=kln(x)⇒y=kxln(x) y=ax+bxln(x) Particular Solution y_p =xz⇒(xz′′+z′)=((log(x))/x^2 ) (xz′)^′ =((log(x))/x^2 )⇒xz′=−((log(x))/x)−(1/x)+c ⇒z=∫(−((log(x))/x^2 )−(1/x^2 ))=(1/x)+((log(x))/x)+(1/x)+cln(x)+d y=2+log(x)+cxln(x)+dx general Solution y=2+log(x)+cxlog(x)+dx,c,d∈R
$$\left({i}\right) \\ $$$${homgenius}\: \\ $$$${x}^{\mathrm{2}} {y}''−{xy}'+{y}=\mathrm{0} \\ $$$$\left.{y}={x}^{{t}} \Rightarrow\left({t}\left({t}−\mathrm{1}\right)−{t}+\mathrm{1}\right)\right){x}^{{t}} =\mathrm{0} \\ $$$$\Rightarrow\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right){x}^{{t}} =\mathrm{0}\Rightarrow{t}=\mathrm{1}\Rightarrow{y}={x} \\ $$$${let}\:{y}={xz}\Rightarrow{y}'={z}+{xz}'\Rightarrow{y}''=\mathrm{2}{z}'+{xz}'' \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left({xz}''+\mathrm{2}{z}'\right)−{x}\left({z}+{xz}'\right)+{xz}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left({xz}''+{z}'\right)=\mathrm{0}\Rightarrow{xz}''+{z}' \\ $$$$\Rightarrow{ln}\left({z}'\right)={ln}\left(\frac{\mathrm{1}}{{x}}\right)+{c}\Rightarrow{z}={kln}\left({x}\right)\Rightarrow{y}={kxln}\left({x}\right) \\ $$$${y}={ax}+{bxln}\left({x}\right) \\ $$$${Particular}\:{Solution} \\ $$$${y}_{{p}} ={xz}\Rightarrow\left({xz}''+{z}'\right)=\frac{{log}\left({x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\left({xz}'\right)^{'} =\frac{{log}\left({x}\right)}{{x}^{\mathrm{2}} }\Rightarrow{xz}'=−\frac{{log}\left({x}\right)}{{x}}−\frac{\mathrm{1}}{{x}}+{c} \\ $$$$\Rightarrow{z}=\int\left(−\frac{{log}\left({x}\right)}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{{x}}+\frac{{log}\left({x}\right)}{{x}}+\frac{\mathrm{1}}{{x}}+{cln}\left({x}\right)+{d} \\ $$$${y}=\mathrm{2}+{log}\left({x}\right)+{cxln}\left({x}\right)+{dx} \\ $$$${general}\:{Solution} \\ $$$${y}=\mathrm{2}+{log}\left({x}\right)+{cxlog}\left({x}\right)+{dx},{c},{d}\in\mathbb{R} \\ $$$$ \\ $$
Commented by niroj last updated on 30/Mar/20
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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