Question Number 90886 by jagoll last updated on 26/Apr/20
$$\frac{{dy}}{{dx}}\:−\frac{\mathrm{4}{y}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{2}}{{x}} \\ $$
Answered by john santu last updated on 26/Apr/20
Commented by john santu last updated on 27/Apr/20
$${typo}\:.\:{y}\:=\:\frac{\int\:{x}^{−\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right){dx}+{C}}{{x}^{−\mathrm{4}} } \\ $$$${y}\:=\:\frac{\int{x}^{−\mathrm{4}} +\mathrm{2}{x}^{−\mathrm{5}} {dx}+{C}}{{x}^{−\mathrm{4}} } \\ $$$${y}=\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\mathrm{4}} +{C}}{{x}^{−\mathrm{4}} } \\ $$$${y}=\:{Cx}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{3}}{x}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by jagoll last updated on 27/Apr/20
$${thank}\:{you}\:{sir}\:{john} \\ $$