Question Number 219651 by SdC355 last updated on 30/Apr/25
$$\mathrm{Solve} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)={e}^{−{kx}} \\ $$
Answered by MrGaster last updated on 30/Apr/25
![Solution of H equation: y_h (x)=C_1 J_ν (x)+C_2 Y_ν (x) Volynski determinant: W(J_ν (x),Y_ν (x))=(2/(πx)) Special solution structure: y_p (x)=−J_ν (x)∫((Y_ν (t)e^(−kt) )/(t^2 ∙(2/(πt))))dt+Y_ν (x)∫((J_ν (t)e^(−kt) )/(t^2 ∙(2/(πt))))dt Simplify the integral term: y_p (x)=(π/2)[Y_ν (x)∫((J_ν (t)e^(−ki) )/t)dt−J_ν (x)∫((Y_ν (t)e^(−kt) )/t)dt] General solution: y(x)=C_1 J_ν (x)+C_2 Y_ν (x)+(π/2)[Y_ν (x)∫((J_ν (t)e^(−ki) )/t)dt−J_ν (x)∫((Y_ν (t)e^(−kt) )/t)dt]](https://www.tinkutara.com/question/Q219666.png)
$$\mathrm{Solution}\:\mathrm{of}\:\mathrm{H}\:\:\mathrm{equation}: \\ $$$${y}_{{h}} \left({x}\right)={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right) \\ $$$$\mathrm{Volynski}\:\mathrm{determinant}: \\ $$$${W}\left({J}_{\nu} \left({x}\right),{Y}_{\nu} \left({x}\right)\right)=\frac{\mathrm{2}}{\pi{x}} \\ $$$$\mathrm{Special}\:\mathrm{solution}\:\mathrm{structure}: \\ $$$${y}_{{p}} \left({x}\right)=−{J}_{\nu} \left({x}\right)\int\frac{{Y}_{\nu} \left({t}\right){e}^{−{kt}} }{{t}^{\mathrm{2}} \centerdot\frac{\mathrm{2}}{\pi{t}}}{dt}+{Y}_{\nu} \left({x}\right)\int\frac{{J}_{\nu} \left({t}\right){e}^{−{kt}} }{{t}^{\mathrm{2}} \centerdot\frac{\mathrm{2}}{\pi{t}}}{dt} \\ $$$$\mathrm{Simplify}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{term}: \\ $$$${y}_{{p}} \left({x}\right)=\frac{\pi}{\mathrm{2}}\left[{Y}_{\nu} \left({x}\right)\int\frac{{J}_{\nu} \left({t}\right){e}^{−{ki}} }{{t}}{dt}−{J}_{\nu} \left({x}\right)\int\frac{{Y}_{\nu} \left({t}\right){e}^{−{kt}} }{{t}}{dt}\right] \\ $$$$\mathrm{General}\:\mathrm{solution}: \\ $$$${y}\left({x}\right)={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right)+\frac{\pi}{\mathrm{2}}\left[{Y}_{\nu} \left({x}\right)\int\frac{{J}_{\nu} \left({t}\right){e}^{−{ki}} }{{t}}{dt}−{J}_{\nu} \left({x}\right)\int\frac{{Y}_{\nu} \left({t}\right){e}^{−{kt}} }{{t}}{dt}\right] \\ $$