Differential problem

**puravida12** · March 28th 2010, 06:04 PM

show that
y = -1/2x . cosx

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

**Prove It** · March 28th 2010, 06:06 PM

Originally Posted by puravida12

show that
y = -1/2x . cosx

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

This is unreadable.

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.

**puravida12** · March 28th 2010, 06:14 PM

-(1/2)x . cosx

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

**Prove It** · March 28th 2010, 06:18 PM

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?

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