# verify that the given function (implicit or explicit) is a solution of the corresponding differential equation

function: $y=ae^x+be^{-x}+x^2$
diff. eqn., $x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 -2 = 0$

Toolbox:
• $\large\frac{d(e^x)}{dx }$$= e^x; \large\frac{d(x)}{dx}$$ = 1$; $\large\frac{d(constant\; term )} {dx}= $$0. The product rule of differentiation states uv = uv' + vu' Step 1: Given :y=ae^x+be^{-x}+x^2 Differentiating the given equation on both sides we get, \large\frac{dy}{dx}$$= ae^x + (-be^-x) + 2x$
Differenting again we get,