CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$y=e^{3x}\sin^{2}x$$ then $$y_{n}=$$


A
3n2e3x13n22e3xcos2x
loader
B
3n2e3x13n22e3xcos[2x+ntan1(3/2)]
loader
C
3n2e3x13n22e3xcos[2x+ntan1(2/3)]
loader
D
3ne3x13n2e3xcos[2x+ntan1(3/2)]
loader

Solution

The correct option is D $$\displaystyle \frac{3^{n}}{2}e^{3x}-\frac{13^{\frac{n}{2}}}{2}e^{3x}\cos[2x+n\tan^{-1}(2/3)]$$
$$y=\dfrac{1}{2}e^{3x}(1-cos 2x)$$ 
$$y=\dfrac{e^{3x}}{2}-\dfrac{e^{3x}cos 2x}{2}$$
$$y'=\dfrac{3e^{3x}}{2}-\dfrac{3e^{3x}cos 2x}{2}+\dfrac{e^{3x}sin 2x}{1}$$
$$y''=\dfrac{3e^{3x}}{2}-\dfrac{\sqrt{13e^{3x}}}{2}\left (\dfrac{-3 cos2x-2 sin2x}{\sqrt{13}} \right )$$
$$y''=\dfrac{3e^{3x}}{2}-\dfrac{13e^{3x}}{2}(cos(2x+tan^{-1}2/3))$$
$$y''=\dfrac{3^{2}e^{3x}}{2}-\dfrac{13}{2}\dfrac{(3e^{3x}cos(2x+tan^{1}2/3)-2e^{3x}sin(2x+tan^{-1}2/3))}{\sqrt{13}}$$
$$y''=\dfrac{3^{2}e^{3x}}{2}-\dfrac{13e^{3x}}{2}cos(2x+tan^{1}2/3)$$
$$y^{n}=\dfrac{3^ne^{3x}}{2}-\dfrac{13^{n/2}}{2}e^{3x}cos(2x+tan^{-1}2/3)$$



Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image