Differential Equations Question 207
Question: The solution of the differential equation $ \frac{dy}{dx}+\frac{3x^{2}}{1+x^{3}}y=\frac{{{\sin }^{2}}x}{1+x^{3}} $ is
Options:
A) $ y(1+x^{3})=x+\frac{1}{2}\sin 2x+c $
B) $ y(1+x^{3})=cx+\frac{1}{2}\sin 2x $
C) $ y(1+x^{3})=cx-\frac{1}{2}\sin 2x $
D) $ y(1+x^{3})=\frac{x}{2}-\frac{1}{4}\sin 2x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{dy}{dx}+\frac{3x^{2}}{1+x^{3}}y=\frac{{{\sin }^{2}}x}{1+x^{3}} $
Here, $ P=\frac{3x^{2}}{1+x^{3}}\Rightarrow $ I.F. $ ={e^{\int _{{}}^{{}}{Pdx}}}={e^{\log (1+x^{3})}}=1+x^{3} $
Thus the solution is $ y.(1+x^{3})=\int _{{}}^{{}}{\frac{{{\sin }^{2}}x}{1+x^{3}}}(1+x^{3})dx=\int _{{}}^{{}}{\frac{1-\cos 2x}{2}}dx $
Therefore $ y(1+x^{3})=\frac{1}{2}x-\frac{\sin 2x}{4}+c $ .
BETA
