Question Number 26240 by sorour87 last updated on 22/Dec/17
$${y}^{\left(\mathrm{2}\right)} −{y}=\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$
Commented by prakash jain last updated on 23/Dec/17
$${y}^{\left(\mathrm{2}\right)} −{y}=\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} }} \\ $$$${y}={y}_{{h}} +{y}_{{p}} \\ $$$${y}''−{y}=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow\lambda=\pm\mathrm{1} \\ $$$${y}_{\mathrm{1}} ={e}^{−{x}} \\ $$$${y}_{\mathrm{2}} ={e}^{{x}} \\ $$$${y}_{{h}} ={c}_{\mathrm{1}} {e}^{{x}} +{c}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)=\begin{vmatrix}{{e}^{−{x}} }&{{e}^{{x}} }\\{\frac{{d}}{{dx}}{e}^{−{x}} }&{\frac{{d}}{{dx}}{e}^{{x}} }\end{vmatrix}=\mathrm{2} \\ $$$${y}_{{p}} =−{y}_{\mathrm{1}} \int\frac{{g}\left({x}\right){y}_{\mathrm{2}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx}+{y}_{\mathrm{2}} \int\frac{{g}\left({x}\right){y}_{\mathrm{1}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx} \\ $$$$\int\frac{{g}\left({x}\right){y}_{\mathrm{2}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx}=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} }}{e}^{{x}} {dx}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left({e}^{{x}} \right) \\ $$$$\int\frac{{g}\left({x}\right){y}_{\mathrm{1}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx}=\int\frac{{e}^{−{x}} }{\:\sqrt{\mathrm{1}−{e}^{\mathrm{2}{z}} }}{dx}=−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}} \sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} } \\ $$$${y}_{{p}} =−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}} \mathrm{sin}^{−\mathrm{1}} {e}^{{x}} −\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−{e}^{\mathrm{2}{z}} } \\ $$
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
$$\boldsymbol{{tnx}} \\ $$