Question Number 91211 by john santu last updated on 28/Apr/20
$${x}\:{dy}\:+\mathrm{5}{y}\:{dx}\:=\:\mathrm{2}{y}^{\mathrm{4}} {x}\:{dx} \\ $$
Commented by john santu last updated on 29/Apr/20
Answered by MWSuSon last updated on 28/Apr/20
$${x}\frac{{dy}}{{dx}}+\mathrm{5}{y}=\mathrm{2}{y}^{\mathrm{4}} {x} \\ $$$${y}^{−\mathrm{4}} \frac{{dy}}{{dx}}+\frac{\mathrm{5}{y}^{−\mathrm{3}} }{{x}}=\mathrm{2} \\ $$$${let}\:{v}={y}^{−\mathrm{3}} \\ $$$$\frac{{dv}}{{dx}}=−\mathrm{3}{y}^{−\mathrm{4}} \frac{{dy}}{{dx}} \\ $$$$\frac{{dv}}{{dx}}−\frac{\mathrm{15}{v}}{{x}}=−\mathrm{6} \\ $$$${I}.{F}={exp}\left(−\mathrm{15}\int\frac{\mathrm{1}}{{x}}{dx}\right)={exp}\left(−\mathrm{15}{log}_{{e}} {x}\right) \\ $$$${x}>\mathrm{0} \\ $$$${x}^{−\mathrm{15}} \frac{{dv}}{{dx}}−\mathrm{15}{x}^{−\mathrm{16}} {v}=\mathrm{6}{x}^{−\mathrm{15}} \\ $$$$\frac{{d}}{{dx}}\left({x}^{−\mathrm{15}} {v}\right)=\mathrm{6}{x}^{−\mathrm{15}} \\ $$$${x}^{−\mathrm{15}} {v}=\int\mathrm{6}{x}^{−\mathrm{15}} {dx} \\ $$$${x}^{−\mathrm{15}} {y}^{−\mathrm{3}} =\frac{\mathrm{3}{x}^{−\mathrm{14}} }{\mathrm{7}}+{c} \\ $$$${y}=^{\mathrm{3}} \sqrt{\frac{{x}^{−\mathrm{15}} }{\frac{\mathrm{3}{x}^{−\mathrm{14}} }{\mathrm{7}}+{c}}} \\ $$