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Question Number 70720 by oyemi kemewari last updated on 07/Oct/19

$${\mathrm{y}}^{″}+\frac{{\mathrm{y}}^{\prime }}{\mathrm{x}}+\frac{4\mathrm{y}}{{\mathrm{x}}^2}=1+{\mathrm{x}}^2$$

Commented by Rasheed.Sindhi last updated on 07/Oct/19

$$\mathcal{T}hankyou!$$

Commented by Rasheed.Sindhi last updated on 07/Oct/19

$$Couldyoupleaseusesmallsizefonts?$$

Commented by oyemi kemewari last updated on 07/Oct/19

ok sir

Commented by mind is power last updated on 07/Oct/19

$$ihavenotfinditinonetry$$$$theideacomesfrom\frac{d}{dx}(y\left(ln\left(x\right)\right)=\frac{y^{\prime }\left(ln\left(x\right)\right)}{x}...$$

Answered by mind is power last updated on 07/Oct/19

$$\iff x^2{y}^{″}+{xy}^{\prime }+4y=x^2(1+x^2)\mathrm{\forall }x\in {\mathbb{R}}^{\ast }...E$$$$lett=ln\left(x\right),y\left(x\right)=g\left(t\right),x=e^t$$$$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=g^{\prime }\left(t\right).{\left(\frac{dx}{dt}\right)}^{-1}=e^{-t}{g}^{\prime }\left(t\right)$$$$\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{dt}{dx}.\frac{d}{dt}\left(e^{-t}{g}^{\prime }\left(t\right)\right)={\left(\frac{dx}{dt}\right)}^{-1}.(-e^{-t}{g}^{\prime }\left(t\right)+e^{-t}{g}^{″}\left(t\right))$$$$=e^{-t}.(-e^{-t}{g}^{\prime }\left(t\right)+e^{-t}{g}^{″}\left(t\right))=-e^{-2t}{g}^{\prime }\left(t\right)+e^{-2t}{g}^{″}\left(t\right)$$$$E\iff e^{2t}(-e^{-2t}{g}^{\prime }\left(t\right)+e^{-2t}{g}^{″}\left(t\right))+e^t\left(e^{-t}{g}^{\prime }\left(t\right)\right)+4g\left(t\right)=e^{2t}(1+e^{2t})$$$$\iff -g^{\prime }\left(t\right)+g^{″}\left(t\right)+g^{\prime }\left(t\right)+4g\left(t\right)=e^{2t}+e^{4t}$$$$\Rightarrow g^{″}\left(t\right)+4g\left(t\right)=e^{2t}+e^{4t}$$$$easytoosolvefromhear$$$$HomogeniousEQ$$$$g^{″}+4g=0\iff g\left(t\right)=acos\left(2t\right)+bsin\left(2t\right),a,b\in \mathbb{R}$$$$particularsolution$$$$let{g}_0={ae}^{2t}+{ce}^{4t}$$$$\Rightarrow 4{ae}^{2t}+16{ce}^{4t}+4{ae}^{2t}+4{ce}^{4t}=e^{2t}+e^{4t}$$$$\Rightarrow a=\frac{1}{8},c=\frac{1}{20}$$$$g\left(t\right)=acos\left(2t\right)+bsin\left(2t\right)+\frac{e^{2t}}{8}+\frac{e^{4t}}{20}$$$$y\left(x\right)=g\left(t\right)=g\left(ln\left(x\right)\right)$$$$Y\left(x\right)=acos\left(2ln\left(x\right)\right)+bsin\left(2ln\left(x\right)\right)+\frac{e^{2ln\left(x\right)}}{8}+\frac{e^{4ln\left(x\right)}}{20}$$$$Y\left(x\right)=acos\left(2ln\left(x\right)\right)+bsin\left(2ln\left(x\right)\right)+\frac{x^2}{8}+\frac{x^4}{20}$$$$$$$$$$

Commented by oyemi kemewari last updated on 07/Oct/19

thank you sir

Commented by mind is power last updated on 07/Oct/19

$$y^{\prime }reWelcom$$

Answered by Joel122 last updated on 08/Oct/19

$$y^{″}+\left(\frac{1}{x}\right)y^{\prime }+\left(\frac{4}{x^2}\right)y=1+x^2$$$$x^2{y}^{″}+{xy}^{\prime }+4y=x^2+x^4...\left(i\right)$$$$$$$$\bullet \mathrm{Homogeneous}\mathrm{solution}$$$$x^2{y}^{″}+{xy}^{\prime }+4y=0\to \mathrm{Euler}-\mathrm{Cauchy}\mathrm{eq}.$$$$\mathrm{Let}y\left(x\right)=x^m\mathrm{be}\mathrm{the}\mathrm{solution}.\mathrm{Then},$$$$x^m[(m^2-m)+m+4]=0$$$$\Rightarrow m^2+4=0\Rightarrow m_1=2i,m_2=-2i$$$$\Rightarrow y_h\left(x\right)=C_1{x}^{2i}+C_2{x}^{-2i}=\alpha \cos \left(\ln x^2\right)+\beta \sin \left(\ln x^2\right)$$$$$$$$\bullet \mathrm{Particular}\mathrm{solution}$$$$\mathrm{Using}\mathrm{method}\mathrm{of}\mathrm{undetermined}\mathrm{parameter},\mathrm{we}\mathrm{have}$$$$y_p\left(x\right)={Ax}^4+{Bx}^3+{Cx}^2+Dx+E$$$$y_p^{{}^{\prime }}\left(x\right)=4{Ax}^3+3{Bx}^2+2Cx+D$$$$y_p^{{}^{″}}\left(x\right)=12{Ax}^2+6Bx+2C$$$$\mathrm{Plugging}\mathrm{these}\mathrm{equations}\mathrm{to}\left(i\right)$$$$x^2[12{Ax}^2+6Bx+2C]+x[4{Ax}^3+3{Bx}^2+2Cx+D]$$$$+4[{Ax}^4+{Bx}^3+{Cx}^2+Dx+E]=x^2+x^4$$$$\Rightarrow 20{Ax}^4+13{Bx}^3+8{Cx}^2+5Dx+4E=x^4+x^2$$$$\mathrm{By}\mathrm{comparing}\mathrm{the}\mathrm{coefficients},\mathrm{we}\mathrm{have}$$$$A=\frac{1}{20},B=0,C=\frac{1}{8},D=0,E=0$$$$\Rightarrow y_p\left(x\right)=\frac{x^4}{20}+\frac{x^2}{8}$$$$$$$$\therefore y\left(x\right)=y_h\left(x\right)+y_p\left(x\right)$$

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