Differential Equation Questions

Question Number 126896 by BHOOPENDRA last updated on 25/Dec/20

$${yy}''−\left({y}'\right)^{\mathrm{2}} ={e}^{{ax}\:} {find}\:{genral}\:{solution}? \\$$

Commented by liberty last updated on 25/Dec/20

$${ok}\:{i}\:{try}\:{solve}\:{it}. \\$$$$\\$$$${for}\:{equation}\:{yy}''−\left({y}'\right)^{\mathrm{2}} ={e}^{{ax}} \\$$$$\:{or}\:{y}''−\frac{\left({y}'\right)^{\mathrm{2}} }{{y}}\:=\:\frac{{e}^{{ax}} }{{y}} \\$$$${put}\:{y}'={v}\left({y}\right)\:;\:{where}\:{y}={f}\left({x}\right) \\$$$${then}\:{y}''\:=\:{v}\left({y}\right){v}'\left({y}\right).\:{the}\:{equation}\:{becomes} \\$$$$\:{v}'−\frac{{v}}{{y}}\:=\:\frac{{e}^{{ax}} }{{y}}.\:\:{the}\:{term}\:{e}^{{ax}} \:{can}\:{be}\:{considered} \\$$$${as}\:{a}\:{constant}.\:{The}\:{integrating}\:{factor}\:{is}\:\frac{\mathrm{1}}{{y}} \\$$$${and}\:{the}\:{solution}\:{given}\:{by}\: \\$$$${v}={y}\left[\:\int\:\frac{{e}^{{ax}} }{{y}^{\mathrm{2}} }\:{dy}\:+\:{b}\:\right]\: \\$$$${v}\:=\:{by}−{e}^{{ax}} \:;\:{give}\:\frac{{dy}}{{dx}}\:−{by}\:=\:−{e}^{{ax}} \\$$$$\:{for}\:{this}\:{equation}\:{integrating}\:{factor} \\$$$${is}\:{e}^{−{bx}} \:{and}\:{the}\:{solution}\:{is} \\$$$${y}\:=\:{e}^{{bx}} \:\left[\:\int\:\left({e}^{−{ax}} \right)\left({e}^{−{bx}} \right){dx}\:+\:{c}\:\right] \\$$$${y}=\:\frac{{e}^{{ax}} }{{b}−{a}}\:+{c}.{e}^{{bx}} \:,\:{b}\neq{a}\:\: \\$$$$\\$$

Commented by mr W last updated on 25/Dec/20

$${how}\:{did}\:{you}\:{get} \\$$$$\:{v}'−\frac{{v}}{{y}}\:=\:\frac{{e}^{{ax}} }{{y}} \\$$$${from} \\$$$${y}''−\frac{\left({y}'\right)^{\mathrm{2}} }{{y}}\:=\:\frac{{e}^{{ax}} }{{y}}? \\$$$$\\$$$${it}\:{should}\:{be} \\$$$$\:{vv}'−\frac{{v}^{\mathrm{2}} }{{y}}\:=\:\frac{{e}^{{ax}} }{{y}} \\$$$$\\$$$${i}\:{don}'{t}\:{think}\:{your}\:{solution}\:{is}\:{correct}. \\$$