A) \[\frac{1}{3}+{{e}^{6}}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{3}+{{e}^{3}}\]
D) \[-\frac{4}{3}\]
Correct Answer: A
Solution :
\[\frac{dy}{dx}+3{{\sec }^{2}}xy=\,{{\sec }^{2}}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\in \left( -\frac{\pi }{3},\,\,\frac{\pi }{3} \right)\] IF \[=\,\,\,{{e}^{\int 3{{\sec }^{2}}xdx}}=e{{\,}^{3\,\,\tan \,x}}\] \[y{{e}^{3\tan x}}=\int{{{e}^{3\,\tan \,x}}\,se{{c}^{2}}xdx}\] \[y{{e}^{3\tan x}}\,=\,\int{{{e}^{t}}\,\frac{dt}{3}}\] put 3 tan \[x=t\text{ }\Rightarrow \text{ }3\,se{{c}^{2}}xdx=dt\] \[y{{e}^{3\tan \,x}}\,=\,\frac{{{e}^{t}}}{3}+C=\,\frac{e{{\,}^{3\tan \,x}}}{3}+C\] \[y\left( \frac{\pi }{4} \right)=\frac{4}{3}\,\,\Rightarrow \,\,\frac{4}{3}{{e}^{3}}\,=\,\frac{{{e}^{3}}}{3}+C\,\,\Rightarrow \,\,C={{e}^{3}}\] \[y\left( \frac{-\pi }{4} \right){{e}^{-3}}=\,\,\frac{{{e}^{-3}}}{3}+{{e}^{3}}\,\Rightarrow \,\,y\left( \frac{-\pi }{4} \right)=\frac{1}{3}+{{e}^{6}}\]
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