Question

If,y=5cosx-3sinx prove that d^2 y/dx^2 +y=0

Answer 1

Given, $\bf{y=5cosx-3sinx}-----(1)$
now, differentiate y with respect to x
dy/dx = d(5cosx - 3sinx)/dx
= d(5cosx)/dx - d(3sinx)/dx
= 5(-sinx) - 3cosx
= -5sinx - 3cosx

so, $\bf{\frac{dy}{dx}=-5sinx-3cosx}$
differentiate dy/dx once again,
$\bf{\frac{d^2y}{dx^2}=\frac{d\{-5sinx-3cos\}}{dx}}\\\\=\bf{-5cosx -3(-sinx)}\\\\=\bf{-(5cosx-3sinx)=-y}$
[ put equation (1), ]

hence, d²y/dx² + y = 0 [ proved /]

Answer 2

y=5cosx-3sinx

differentiating wrt x we get...

$\frac{dy}{dx} = - 5 \sin(x) - 3 \cos(x) \\ \frac{ {d}^{2}y }{d {x}^{2} } = - 5cos(x) + 3 \sin(x) \\ now \: taking \: lhs \\ - 5 \cos(x) + 3sin(x) + y \\ - 5 \cos(x) + 3 \sin(x) + (5 \cos(x) - 3 \sin(x) ) \\ 0 = rhs$