$\cos\left(\frac{\pi}{4}-x\right)-sin\left(\frac{\pi}{4}+x\right)$

$x'=bx$

$-14\:\left(-5\right)\:+\:5\:$

$\left(1\right)\left(5\right)-0.5\left(5.\right)$

$tx'-2x=2t^4\:,x\left(1\right)=2$

$\frac{dy}{dx}=\frac{y\left(2-y\right)}{\left(1+x^2\right)}$

$8x^3y^3\left(10x^5\right)$