# Given that y=(k+x)cosx where k is a constant find d^2y\dx^2 +y and show that it's independent of k ?

Feb 15, 2017

See below.

#### Explanation:

$y = \left(k + x\right) \cos x$

By the Product Rule:

$y ' = \cos x - \left(k + x\right) \sin x$

Same again:

$y ' ' = - \sin x - \sin x - \left(k + x\right) \cos x$

$= - 2 \sin x - \left(k + x\right) \cos x$

$\implies y ' ' + y = - 2 \sin x$

This is independent of k.

To understand why, consider the simpler: $y = k \cos x$.

In this case:
$y ' = - k \sin x$
$y ' ' = - k \cos x$
$\implies y ' ' + y = 0$

To look at it another way, $y = k \cos x$ is one of the (complementary/null) solutions to the homogeneous linear DE $y ' ' + y = 0$. The other is $y = {k}_{2} \sin x$. Try that too and see.