# GATE2014-1-46

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The function $f(x) =x \sin x$ satisfies the following equation: $$f''(x) + f(x) +t \cos x = 0$$. The value of $t$ is______.
in Calculus
retagged

$f'(x) =x\cos(x) + \sin(x)$

$f''(x)=x(-\sin x) +\cos x +\cos x$

now $f''(x)+f(x)+t \cos x= 0$

$\Rightarrow x(-\sin x)+\cos x+\cos x+x\sin x+t\cos x=0$

$\Rightarrow 2\cos x+t\cos x = 0$
$\Rightarrow \cos x(t+2)=0$
$\Rightarrow t+2=0, t=-2$

edited

Hence, t = -2

t=-2
0
how?
1 vote
We have f(x) = x sin x

β f'(x) = x cos x + sin x

β fβ²β²(x) = x (β sin x ) + cos x + cos x = (βx sin x ) + 2 cos x

Now, it is given that f(x) = x sin x satisfies the equation fβ²β²(x) + f(x) + t cos x = 0

β (βx sin x ) + 2 cos x + x sin x + t cos x = 0

β 2 cos x + t cos x = 0

β cos x ( t + 2 ) = 0

β t + 2 = 0

β t = β2

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