A) \[\frac{1}{3}+{{e}^{6}}\]
B) \[\frac{1}{3}\]
C) \[-\frac{4}{3}\]
D) \[\frac{1}{3}+{{e}^{3}}\]
Correct Answer: A
Solution :
| \[\frac{dy}{dx}+(3{{\sec }^{3}}x)y={{\sec }^{2}}x\] |
| This linear differential equation |
| Integrating factor \[={{e}^{\int{3{{\sec }^{2}}xdx}}}={{e}^{3\tan x}}\] |
| Hence, \[y.{{e}^{3\tan x}}={{e}^{\int{3\tan x}}}.{{\sec }^{2}}xdx\] |
| \[\Rightarrow \] \[y.{{e}^{3\tan x}}=\frac{{{e}^{3\tan x}}}{3}+c\] |
| \[\Rightarrow \] \[y=C{{e}^{-3\tan x}}+\frac{1}{3}\] |
| Given, \[y\left( \frac{\pi }{4} \right)=\frac{4}{3}\] |
| \[\Rightarrow \] \[\frac{4}{3}=C{{e}^{-3}}+\frac{1}{3}\] |
| \[\Rightarrow \] \[C={{e}^{3}}\] |
| Hence, \[y\left( \frac{\pi }{4} \right)={{e}^{3}}.{{e}^{3}}+\frac{1}{3}={{e}^{6}}+\frac{1}{3}.\] |
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