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Question Number 103037 by bramlex last updated on 12/Jul/20

$${4\mathrm{x}}^2{\mathrm{y}}^{″}+\mathrm{y}=0,\mathrm{x}>0$$

Answered by maths mind last updated on 12/Jul/20

$$lety=x^t$$ $$\Rightarrow 4x^2\left(t(t-1)\right)x^{t-2}+x^t=0$$ $$\iff (4t^2-4t+1)x^t=0$$ $$\iff {(2t-1)}^2{x}^t=0\Rightarrow t=\frac{1}{2}$$ $$lety=\sqrt{x}z$$ $$\Rightarrow y^{\prime }=z^{\prime }\sqrt{x}+\frac{z}{2\sqrt{x}},y^{″}=\frac{z^{\prime }}{\sqrt{x}}+z^{″}\sqrt{x}-\frac{z}{4x\sqrt{x}}$$ $$\iff 4x\sqrt{x}{z}^{\prime }+4x^2\sqrt{x}{z}^{″}-z\sqrt{x}+z\sqrt{x}=0$$ $$\Rightarrow z^{\prime }+{xz}^{″}=0$$ $$\Rightarrow \frac{z^{″}}{z^{\prime }}=-\frac{1}{x}\Rightarrow ln\left(z^{\prime }\right)=-ln\left(x\right)+c$$ $$\Rightarrow z^{\prime }=\frac{k}{x}\Rightarrow z=kln\left(x\right)+c$$ $$\Rightarrow y=(kln\left(x\right)+c)\sqrt{x}$$

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