Differential Equations Question 250
Question: The solution of the differential equation $ \frac{dy}{dx}+\frac{2yx}{1+x^{2}}=\frac{1}{{{(1+x^{2})}^{2}}} $ is:
Options:
A) $ y(1+x^{2})=c+{{\tan }^{-1}}x $
B) $ \frac{y}{1+x^{2}}=c+{{\tan }^{-1}}x $
C) $ y\log (1+x^{2})=c+{{\tan }^{-1}}x $
D) $ y(1+x^{2})=c+{{\sin }^{-1}}x $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given differential equation is $ \frac{dy}{dx}+\frac{2yx}{1+x^{2}}=\frac{1}{{{(1+x^{2})}^{2}}} $ which is a linear. Differential equation of the form: $ \frac{dy}{dx}+Py=Q $ On comparing, we have $ P=\frac{2x}{1+x^{2}} $ and $ Q=\frac{1}{{{(1+x^{2})}^{2}}} $ I.F $ ={e^{\int{2x/1+x^{2}dx}}}={e^{\log (1+x^{2})}}=(1+x^{2}) $
$ \therefore $ $ Solutions $ y(1+x^{2}) $ $ =\int{\frac{1}{{{(1+x^{2})}^{2}}}(1+x^{2})dx+c} $
$ \Rightarrow y(1+x^{2})=\int{\frac{1}{(1+x^{2})}dx+c} $
$ \Rightarrow y(1+x^{2})={{\tan }^{-1}}x+c $
BETA
