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Question Number 95801 by bobhans last updated on 27/May/20

$${2\mathrm{y}}^{″}-{\mathrm{y}}^{{}^{\prime }}=1;\mathrm{y}\left(0\right)=0;{\mathrm{y}}^{\prime }\left(0\right)=1$$

Answered by mr W last updated on 27/May/20

$$u=y^{\prime }$$$$y^{″}=\frac{du}{dx}$$$$2\frac{du}{dx}-u=1$$$$\frac{2du}{u+1}=dx$$$$2\ln (u+1)=x+C_1$$$$2\ln (1+1)=C_1$$$$\ln \frac{u+1}{2}=\frac{x}{2}$$$$u=\frac{dy}{dx}=2e^{\frac{x}{2}}-1$$$$y=\int (2e^{\frac{x}{2}}-1)dx$$$$y=4e^{\frac{x}{2}}-x+C_2$$$$0=4+C_2$$$$\Rightarrow y=4(e^{\frac{x}{2}}-1)-x$$

Answered by john santu last updated on 28/May/20

$$\mathrm{auxilary}\mathrm{equation}$$$$2{\lambda }^2-\lambda =0$$$$\lambda (2\lambda -1)=0;\lambda =0,\frac{1}{2}$$$${\mathrm{y}}_{\mathrm{h}}=\mathrm{A}+{\mathrm{Be}}^{\frac{1}{2}\mathrm{x}}$$$$\mathrm{particular}\Rightarrow {\mathrm{y}}_{\mathrm{p}}=-\mathrm{x}$$$$\mathrm{complete}\mathrm{solution}$$$$\mathrm{y}=\mathrm{A}+{\mathrm{Be}}^{\frac{1}{2}\mathrm{x}}-\mathrm{x}$$$${\mathrm{y}}^{\prime }=\frac{1}{2}{\mathrm{Be}}^{\frac{1}{2}\mathrm{x}}-1\Rightarrow {\mathrm{y}}^{\prime }\left(0\right)=1$$$$1=\frac{1}{2}\mathrm{B}-1\Rightarrow \mathrm{B}=4$$$$\mathrm{y}=\mathrm{A}+{4\mathrm{e}}^{\frac{1}{2}\mathrm{x}}-\mathrm{x}\Rightarrow \mathrm{y}\left(0\right)=0$$$$0=\mathrm{A}+4-0\Rightarrow \mathrm{A}=-4$$$$\therefore \mathrm{y}={4\mathrm{e}}^{\frac{1}{2}x}-4-x$$

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