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Question Number 50135 by peter frank last updated on 14/Dec/18

Commented by maxmathsup by imad last updated on 01/Jan/19

$$letsolve(2x-1)y^{{}^{″}}-y^{{}^{\prime }}=0withy\left(0\right)=2and{y}^{{}^{\prime }}\left(0\right)=3letusechangement$$$$y^{{}^{\prime }}=Z\Rightarrow (2x-1)z^{{}^{\prime }}-z=0\Rightarrow (2x-1)z^{{}^{\prime }}=z\Rightarrow \frac{z^{{}^{\prime }}}{z}=\frac{1}{2x-1}\Rightarrow ln\mid z\mid =\frac{1}{2}ln\mid 2x-1\mid +k\Rightarrow$$$$\Rightarrow z=K\sqrt{\mid 2x-1\mid }lettakex<\frac{1}{2}\Rightarrow z=K\sqrt{-2x+1}\Rightarrow y^{{}^{\prime }}=k\sqrt{-2x+1}\Rightarrow$$$$y=k\int \sqrt{-2x+1}dx+c_0=k\int {(-2x+1)}^{\frac{1}{2}}dx+c_0$$$$=k\frac{2}{3}(-\frac{1}{2}){(-2x+1)}^{\frac{3}{2}}+c_0=c_0-\frac{k}{3}{(-2x+1)}^{\frac{3}{2}}\Rightarrow$$$$y\left(x\right)=c_0-\frac{k}{3}(-2x+1)\sqrt{-2x+1}$$$$y\left(0\right)=2\Rightarrow c_0-\frac{k}{3}=2\Rightarrow 3c_0-k=6$$$$y^{{}^{\prime }}\left(0\right)=3\Rightarrow z\left(0\right)=3\Rightarrow k=3\Rightarrow 3c_0=9\Rightarrow c_0=3\Rightarrow y\left(x\right)=3-(-2x+1)\sqrt{-2x+1}$$$$\Rightarrow y\left(x\right)=3+(2x-1)\sqrt{-2x+1}thisisthesolutionon]-\mathrm{\infty },\frac{1}{2}[.$$$$z$$$$$$

Commented by maxmathsup by imad last updated on 01/Jan/19

$$sorryihavesolved(2x-1)y^{{}^{″}}-y^{{}^{\prime }}=0not(2x-1)y^{{}^{″}}-2y=0butthewayis$$$$thesame.$$

Answered by peter frank last updated on 14/Dec/18

$$(2x-1)\frac{d^{2}x}{{dx}^2}-2\frac{dy}{dx}=0$$$$letp=\frac{d^{2}y}{{dx}^2}\frac{dp}{dx}=\frac{d^{2}y}{{dx}^2}$$$$(2x-1)\frac{dp}{dx}-2p=0$$$$\int \frac{dp}{p}=\int \frac{\mathrm{dx}}{2\mathrm{x}-1}$$$$lnp=ln(2x-1)+B$$$$lnp=ln(2x-1)+\ln B$$$$p=B(2x-1)$$$$but$$$$fromp=\frac{dy}{dx}\frac{dp}{dx}=\frac{d^{2}y}{{dx}^2}$$$$p=\frac{dy}{dx}=B(2x-1)$$$$given\frac{dy}{dx}=3x=0\mathrm{B}=-3$$$$\frac{dy}{dx}=B(2x-1)$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-3(2\mathrm{x}-1)$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-6\mathrm{x}+3$$$$\int \mathrm{dy}=\int (-6\mathrm{x}+3)\mathrm{dx}$$$$\mathrm{y}=-3\mathrm{x}+3\mathrm{x}+\mathrm{D}$$$$\mathrm{given}\mathrm{x}=0\mathrm{y}=2$$$$\mathrm{D}=2$$$$\mathrm{y}=-3\mathrm{x}+3\mathrm{x}+2$$$$\mathrm{required}\mathrm{solution}$$$$$$

Answered by peter frank last updated on 14/Dec/18

$$\mathrm{x}(\mathrm{x}+\mathrm{y})\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2+\mathrm{xy}-{3\mathrm{y}}^2$$$$(1+\frac{\mathrm{y}}{\mathrm{x}})\frac{\mathrm{dy}}{\mathrm{dx}}=1+\frac{\mathrm{y}}{\mathrm{x}}-3{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2$$$$\mathrm{from}\mathrm{y}=\mathrm{vx}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}$$$$(1+\mathrm{v})(\mathrm{v}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}})=1+\mathrm{v}-{3\mathrm{v}}^2$$$$\frac{(1+\mathrm{v})}{1-{4\mathrm{v}}^2}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}$$$$\int \frac{\mathrm{vdv}}{1-{4\mathrm{v}}^2}+\int \frac{\mathrm{dv}}{1-{4\mathrm{v}}^2}=\ln \mathrm{x}+\mathrm{A}$$$$-\frac{1}{8}\int \frac{8\mathrm{v}}{1-{4\mathrm{v}}^2}\mathrm{dv}+\frac{1}{2}\int \frac{\mathrm{dv}}{1-2\mathrm{v}}+\frac{1}{2}\int \frac{\mathrm{dv}}{1-2\mathrm{v}}=\mathrm{lnx}+\mathrm{A}$$$$-\ln (1-{4\mathrm{v}}^2)-2\ln (1-2\mathrm{v})+2\ln (1+2\mathrm{v})=8\ln \left(\mathrm{Ax}\right)$$$$\ln \left[\frac{{(1+2\mathrm{v})}^2}{{(1-2\mathrm{v})}^2(1-{4\mathrm{v}}^2)}\right]=\ln {\mathrm{A}}^8{\mathrm{x}}^8$$$$\frac{1+2\mathrm{v}}{{1-2\mathrm{v})}^3}={\mathrm{A}}^8{\mathrm{x}}^8{\mathrm{A}}^8=\mathrm{C}$$$$\mathrm{but}\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}\mathrm{and}\mathrm{simplify}$$$$\mathrm{x}+2\mathrm{y}={\mathrm{Cx}}^6{(\mathrm{x}-2\mathrm{y})}^3$$$$\mathrm{Required}\mathrm{solution}$$$$$$

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