The functions $u=e^x \sin x ; v=e^x \cos x$ satisfy the equation

all the above

We have,

$u=e^x \sin x \Rightarrow \frac{d u}{d x}=e^x \sin x+e^x \cos x=u+v$

$v=e^x \cos x \Rightarrow \frac{d v}{d x}=e^x \cos x+e^x \sin x=v-u$

∴   $v \frac{d u}{d x}-\frac{u d v}{d x}=v(u+v)-u(v-u)=u^2+v^2$

$\frac{d^2 u}{d x^2}=\frac{d u}{d x}+\frac{d v}{d x}=u+v+v-u=2 v$

and, $\frac{d^2 v}{d x^2}=\frac{d v}{d x}-\frac{d u}{d x}=(v-u)-(v+u)=-2 u$.