solve-y-y-y-

Question Number 68133 by mr W last updated on 05/Sep/19
$${solve}\:{y}''={y}'{y} \\$$
Answered by mind is power last updated on 05/Sep/19
$$\Rightarrow{y}^{'} =\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{s} \\$$$${solve}\:{this}\:{we}\:{need}\:{value}\left[{of}\:{f}'\left(\mathrm{0}\right)\right. \\$$$${if}\:{f}'\left(\mathrm{0}\right)={a}>\mathrm{0} \\$$$$\Rightarrow\frac{{y}^{'} }{\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{a}}=\mathrm{1}\Rightarrow\sqrt{\frac{\mathrm{2}}{{a}}}{arctan}\left(\frac{{y}}{\mathrm{2}}\right)={x}+{s} \\$$$${y}=\mathrm{2}{tg}\left(\frac{{x}+{s}}{\:\sqrt{\frac{\mathrm{2}}{{a}}}}\right) \\$$$$\\$$$$\\$$
Commented by mr W last updated on 06/Sep/19
$${thanks}\:{sir}! \\$$
Answered by Kunal12588 last updated on 06/Sep/19
$$\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{dy}}{{dx}}×{y} \\$$$$\Rightarrow\frac{{dy}}{{dx}}=\int{ydy} \\$$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{c}\Rightarrow\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} +\mathrm{2}{c}}{\mathrm{2}} \\$$$$\Rightarrow{x}=\mathrm{2}\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{2}{c}} \\$$$$\Rightarrow{x}=\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{c}}}×{tan}^{−\mathrm{1}} \left(\frac{{y}}{\:\sqrt{\mathrm{2}{c}}}\right)+{k} \\$$$$\Rightarrow{x}+{k}=\sqrt{\frac{\mathrm{2}}{{c}}}{tan}^{−\mathrm{1}} \left(\frac{{y}}{\:\sqrt{\mathrm{2}{c}}}\right) \\$$$$\Rightarrow\frac{{y}}{\:\sqrt{\mathrm{2}{c}}}={tan}\left(\left({x}+{k}\right)\sqrt{\frac{{c}}{\mathrm{2}}}\right) \\$$$$\Rightarrow{y}=\sqrt{\mathrm{2}{c}}\:{tan}\left(\left({x}+{k}\right)\sqrt{\frac{{c}}{\mathrm{2}}}\right) \\$$
Commented by Kunal12588 last updated on 06/Sep/19
$${is}\:{this}\:{correct}…\:{this}\:{is}\:{my}\:{first}\:{ever}\:{encounter} \\$$$${of}\:{differential}\:{equation} \\$$
Commented by mr W last updated on 06/Sep/19
$${it}'{s}\:{basically}\:{right},\:{but}\:{you}\:{have} \\$$$${assumed}\:{that}\:{c}>\mathrm{0}.\:{if}\:{c}<\mathrm{0},\:{the} \\$$$${result}\:{is}\:{different}. \\$$
Commented by Kunal12588 last updated on 07/Sep/19
$${if}\:{c}=\mathrm{0} \\$$$${x}=\mathrm{2}\int\frac{{dy}}{{y}^{\mathrm{2}} }=\mathrm{2}\frac{−\mathrm{1}}{{y}}+{k} \\$$$$\Rightarrow\frac{\mathrm{1}}{{k}−{x}}=\left(\frac{{y}}{\mathrm{2}}\right) \\$$$${y}=\frac{\mathrm{2}}{{k}−{x}} \\$$$${if}\:{c}<\mathrm{0} \\$$$${x}=\mathrm{2}\int\frac{{dy}}{{y}^{\mathrm{2}} −\mathrm{2}{c}} \\$$$$\Rightarrow{x}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{c}}}{log}\left(\frac{{y}−\sqrt{\mathrm{2}{c}}}{{y}+\sqrt{\mathrm{2}{c}}}\right)+{k} \\$$$$\Rightarrow\left({x}−{k}\right)\sqrt{\mathrm{2}{c}}={log}\left(\frac{{y}−\sqrt{\mathrm{2}{c}}}{{y}+\sqrt{\mathrm{2}{c}}}\right) \\$$$${what}\:{should}\:{i}\:{do}\:{now} \\$$