Question
What is the solution of the differential equation $$\frac{{dx}}{{dy}} + \frac{x}{y} - {y^2} = 0\,?$$
where $$c$$ is an arbitrary constant
A.
$$xy = {x^4} + c$$
B.
$$xy = {y^4} + c$$
C.
$$4xy = {y^4} + c$$
D.
$$3xy = {y^3} + c$$
Answer :
$$4xy = {y^4} + c$$
Solution :
$$\frac{{dx}}{{dy}} + \frac{x}{y} - {y^2} = 0\,;\,\frac{{dx}}{{dy}} + \frac{x}{y} = {y^2}$$
This is a linear differential equation of the form
$$\eqalign{
& \frac{{dx}}{{dy}} + {P_1}x = {Q_1}\,; \cr
& {\text{Here, }}P = \frac{1}{y}{\text{ and }}Q = {y^2} \cr
& \therefore {\text{ I}}{\text{.F}}{\text{.}} = {e^{\int {P\,dy} }} = {e^{\int {\frac{1}{y}\,dy} }} = {e^{\log \,y}} = y \cr} $$
So, required solution is
$$\eqalign{
& x.y = \int {{y^2}.y\,dy + c\,;\,xy} = \int {{y^3}dy + c} \cr
& xy = \frac{{{y^4}}}{4} + c\,;\,4xy = {y^4} + c \cr} $$