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Question Number 86640 by niroj last updated on 29/Mar/20

$$\mathbf{Solve}\mathbf{the}\mathbf{differential}\mathbf{equations}:$$$$\left(\mathbf{i}\right).{\mathbf{x}}^2\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}-\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}+\mathbf{y}=\mathbf{log}\mathbf{x}.$$$$\left(\mathbf{ii}\right).{(\mathbf{x}+2)}^2\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}-4(\mathbf{x}+2)\frac{\mathbf{dy}}{\mathbf{dx}}+6\mathbf{y}=\mathbf{x}.$$$$$$

Answered by mind is power last updated on 30/Mar/20

$$\left(i\right)$$$$homgenius$$$$x^2{y}^{″}-{xy}^{\prime }+y=0$$$$y=x^t\Rightarrow (t(t-1)-t+1))x^t=0$$$$\Rightarrow (t^2-2t+1)x^t=0\Rightarrow t=1\Rightarrow y=x$$$$lety=xz\Rightarrow y^{\prime }=z+{xz}^{\prime }\Rightarrow y^{″}=2z^{\prime }+{xz}^{″}$$$$\Rightarrow x^2({xz}^{″}+2z^{\prime })-x(z+{xz}^{\prime })+xz=0$$$$\Rightarrow x^2({xz}^{″}+z^{\prime })=0\Rightarrow {xz}^{″}+z^{\prime }$$$$\Rightarrow ln\left(z^{\prime }\right)=ln\left(\frac{1}{x}\right)+c\Rightarrow z=kln\left(x\right)\Rightarrow y=kxln\left(x\right)$$$$y=ax+bxln\left(x\right)$$$$ParticularSolution$$$$y_p=xz\Rightarrow ({xz}^{″}+z^{\prime })=\frac{log\left(x\right)}{x^2}$$$${\left({xz}^{\prime }\right)}^{{}^{\prime }}=\frac{log\left(x\right)}{x^2}\Rightarrow {xz}^{\prime }=-\frac{log\left(x\right)}{x}-\frac{1}{x}+c$$$$\Rightarrow z=\int (-\frac{log\left(x\right)}{x^2}-\frac{1}{x^2})=\frac{1}{x}+\frac{log\left(x\right)}{x}+\frac{1}{x}+cln\left(x\right)+d$$$$y=2+log\left(x\right)+cxln\left(x\right)+dx$$$$generalSolution$$$$y=2+log\left(x\right)+cxlog\left(x\right)+dx,c,d\in \mathbb{R}$$$$$$

Commented by niroj last updated on 30/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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