Differential Equations Question 315
Question: What is the solution of the differential equation $ \frac{dx}{dy}+\frac{x}{y}-y^{2}=0 $ -
Options:
A) $ xy=x^{4}+c $
B) $ xy=y^{4}+c $
C) $ 4xy=y^{4}+c $
D) $ 3xy=y^{3}+c $ where c is an arbitrary constant.
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \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 $ \frac{dx}{dy}+P_1x=Q_1; $ Here, $ P=\frac{1}{y} $ and $ Q=y^{2} $
$ \therefore $ I.F. $ ={e^{\int{Pdy}}}={e^{\int{\frac{1}{y}dy}}}={e^{\log y}}=y $ So, required solution is $ x\cdot y=\int{y^{2}\cdot ydy+c;xy=\int{y^{3}dy+c}} $
$ xy=\frac{y^{4}}{4}+c;4xy=y^{4}+c $
BETA
