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Question Number 156931 by cortano last updated on 17/Oct/21

$$(1-x^2)y^{″}-4{xy}^{\prime }-(1+x^2)y=0$$

Commented by john_santu last updated on 17/Oct/21

$$y=\frac{\sin x}{1-x^2}+c$$

Commented by aliyn last updated on 18/Oct/21

$$$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$$$$${\mathit{y}}^{{}^{\prime }}=\frac{4\mathit{x}\pm \sqrt{16{\mathit{x}}^2+4+4{\mathit{x}}^2}}{2(1-{\mathit{x}}^2)}$$$$$$$${\mathit{y}}^{{}^{\prime }}=\frac{2\mathit{x}\pm \sqrt{5{\mathit{x}}^2+1}}{1-{\mathit{x}}^2}$$$$$$$$\mathit{d}\mathit{y}=\frac{2\mathit{x}}{1-{\mathit{x}}^2}\pm \frac{\sqrt{5{\mathit{x}}^2+1}}{1-{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$$$$$\mathit{y}=\mathit{l}\mathit{n}\mid \frac{1}{\sqrt{1-{\mathit{x}}^2}}\mid \pm \int \frac{\sqrt{5{\mathit{x}}^2+1}}{1-{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$$$$$\mathit{n}\mathit{o}\mathit{w}:\mathit{x}=\frac{1}{\sqrt{5}}\mathit{t}\mathit{a}\mathit{n}\mathit{y}\to \mathit{d}\mathit{x}=\frac{1}{\sqrt{5}}{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{y}\mathit{d}\mathit{y}$$$$$$$$\mathit{c}\mathit{o}\mathit{m}\mathit{p}\mathit{l}\mathit{e}\mathit{t}\mathit{e}$$

Answered by mindispower last updated on 17/Oct/21

$$z=(1-x^2)y$$$$z^{\prime }=-2xy+(1-x^2)y^{\prime }$$$$z^{″}=-2y-4{xy}^{\prime }+(1-x^2)y^{″}$$$$z+z^{″}=-(1+x^2)y-4{xy}^{\prime }+(1-x^2)y^{″}$$$$z+z^{″}=0\Rightarrow z=acos\left(x\right)+bsin\left(x\right)$$$$y=\frac{acos\left(x\right)+bsin\left(x\right)}{1-x^2}$$$$$$

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