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

y^((2)) −y=(1−e^(2x) )^((−1)/2)

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

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

y^((2)) −y=(1−e^(2x) )^((−1)/2)   g(x)=(1/(√(1−e^(2x) )))  y=y_h +y_p   y′′−y=0  λ^2 −1=0⇒λ=±1  y_1 =e^(−x)   y_2 =e^x   y_h =c_1 e^x +c_2 e^(−x)   W(y_1 ,y_2 )= determinant ((e^(−x) ,e^x ),(((d/dx)e^(−x) ),((d/dx)e^x )))=2  y_p =−y_1 ∫((g(x)y_2 )/(W(y_1 ,y_2 )))dx+y_2 ∫((g(x)y_1 )/(W(y_1 ,y_2 )))dx  ∫((g(x)y_2 )/(W(y_1 ,y_2 )))dx=∫(1/(√(1−e^(2x) )))e^x dx=−(1/2)sin^(−1) (e^x )  ∫((g(x)y_1 )/(W(y_1 ,y_2 )))dx=∫(e^(−x) /(√(1−e^(2z) )))dx=−(1/2)e^(−x) (√(1−e^(2x) ))  y_p =−(1/2)e^(−x) sin^(−1) e^x −(1/2)(√(1−e^(2z) ))

$${y}^{\left(\mathrm{2}\right)} −{y}=\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} }} \\ $$$${y}={y}_{{h}} +{y}_{{p}} \\ $$$${y}''−{y}=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow\lambda=\pm\mathrm{1} \\ $$$${y}_{\mathrm{1}} ={e}^{−{x}} \\ $$$${y}_{\mathrm{2}} ={e}^{{x}} \\ $$$${y}_{{h}} ={c}_{\mathrm{1}} {e}^{{x}} +{c}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)=\begin{vmatrix}{{e}^{−{x}} }&{{e}^{{x}} }\\{\frac{{d}}{{dx}}{e}^{−{x}} }&{\frac{{d}}{{dx}}{e}^{{x}} }\end{vmatrix}=\mathrm{2} \\ $$$${y}_{{p}} =−{y}_{\mathrm{1}} \int\frac{{g}\left({x}\right){y}_{\mathrm{2}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx}+{y}_{\mathrm{2}} \int\frac{{g}\left({x}\right){y}_{\mathrm{1}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx} \\ $$$$\int\frac{{g}\left({x}\right){y}_{\mathrm{2}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx}=\int\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} }}{e}^{{x}} {dx}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left({e}^{{x}} \right) \\ $$$$\int\frac{{g}\left({x}\right){y}_{\mathrm{1}} }{\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)}{dx}=\int\frac{{e}^{−{x}} }{\sqrt{\mathrm{1}−{e}^{\mathrm{2}{z}} }}{dx}=−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}} \sqrt{\mathrm{1}−{e}^{\mathrm{2}{x}} } \\ $$$${y}_{{p}} =−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}} \mathrm{sin}^{−\mathrm{1}} {e}^{{x}} −\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−{e}^{\mathrm{2}{z}} } \\ $$

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

y=y_h +y_p   y_h  and y_p  given in comments.

$${y}={y}_{{h}} +{y}_{{p}} \\ $$$${y}_{{h}} \:\mathrm{and}\:{y}_{{p}} \:\mathrm{given}\:\mathrm{in}\:\mathrm{comments}. \\ $$

Commented by sorour87 last updated on 23/Dec/17

tnx

$$\boldsymbol{{tnx}} \\ $$

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