# Math Help - Differential problem

1. ## Differential problem

show that
y = -1/2x . cosx

is/isn't a solution to y" + y = sinx

2. Originally Posted by puravida12
show that
y = -1/2x . cosx

is/isn't a solution to y" + y = sinx

Is it $-\frac{1}{2}x\cos{x}$, or is it $-\frac{1}{2x}\cos{x}$, or is it $-\frac{1}{2x\cos{x}}$?

Please use LaTeX or put brackets where they're needed.

3. -(1/2)x . cosx

I apologize, it has been a long night of calculus.

4. If it's a solution, then

$y = -\frac{1}{2}x\cos{x}$

$y' = -\frac{1}{2}(\cos{x} - x\sin{x})$

$= \frac{1}{2}x\sin{x} - \frac{1}{2}\cos{x}$.

$y'' = \frac{1}{2}(\sin{x} + x\cos{x}) + \frac{1}{2}\sin{x}$

$= \frac{1}{2}\sin{x} + \frac{1}{2}x\cos{x} + \frac{1}{2}\sin{x}$

$= \frac{1}{2}x\cos{x} + \sin{x}$.

So $y'' + y = \frac{1}{2}x\cos{x} + \sin{x} - \frac{1}{2}x\cos{x}$

$= \sin{x}$.

Do you think it's a solution?