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Question Number 26240 by sorour87 last updated on 22/Dec/17

$$y^{\left(2\right)}-y={(1-e^{2x})}^{\frac{-1}{2}}$$

Commented by prakash jain last updated on 23/Dec/17

$$y^{\left(2\right)}-y={(1-e^{2x})}^{\frac{-1}{2}}$$$$g\left(x\right)=\frac{1}{\sqrt{1-e^{2x}}}$$$$y=y_h+y_p$$$$y^{″}-y=0$$$${\lambda }^2-1=0\Rightarrow \lambda =\pm 1$$$$y_1=e^{-x}$$$$y_2=e^x$$$$y_h=c_1{e}^x+c_2{e}^{-x}$$$$\mathcal{W}(y_1,y_2)=\left|\begin{array}{cc}{e}^{-x}& e^x\\ \frac{d}{dx}{e}^{-x}& \frac{d}{dx}{e}^x\end{array}\right|=2$$$$y_p=-y_1\int \frac{g\left(x\right)y_2}{\mathcal{W}(y_1,y_2)}dx+y_2\int \frac{g\left(x\right)y_1}{\mathcal{W}(y_1,y_2)}dx$$$$\int \frac{g\left(x\right)y_2}{\mathcal{W}(y_1,y_2)}dx=\int \frac{1}{\sqrt{1-e^{2x}}}{e}^{x}dx=-\frac{1}{2}{\sin }^{-1}\left(e^x\right)$$$$\int \frac{g\left(x\right)y_1}{\mathcal{W}(y_1,y_2)}dx=\int \frac{e^{-x}}{\sqrt{1-e^{2z}}}dx=-\frac{1}{2}{e}^{-x}\sqrt{1-e^{2x}}$$$$y_p=-\frac{1}{2}{e}^{-x}{\sin }^{-1}{e}^x-\frac{1}{2}\sqrt{1-e^{2z}}$$

Answered by prakash jain last updated on 23/Dec/17

$$y=y_h+y_p$$$$y_h\mathrm{and}{y}_p\mathrm{given}\mathrm{in}\mathrm{comments}.$$

Commented by sorour87 last updated on 23/Dec/17

$$\mathit{t}\mathit{n}\mathit{x}$$

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