Differential Equations Question 214
Question: The solution of the equation $ x\frac{dy}{dx}+3y=x $ is
Options:
A) $ x^{3}y+\frac{x^{4}}{4}+c=0 $
B) $ x^{3}y=\frac{x^{4}}{4}+c $
C) $ x^{3}y+\frac{x^{4}}{4}=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ x\frac{dy}{dx}+3y=x $
Therefore $ \frac{dy}{dx}+\frac{3y}{x}=1 $
It is in the form of $ \frac{dy}{dx}+Py=Q $
So, I.F. $ ={e^{\int _{{}}^{{}}{Pdx}}}=e^{3^{\int _{{}}^{{}}{\frac{1}{x}dx}}}={e^{3\log x}}=x^{3} $
Hence required solution is $ y+x^{2}+2x+2=ce^{x} $
Therefore $ yx^{3}=\frac{x^{4}}{4}+c $ .
BETA
