Question

If y=Ae^mx +Be^nx show that d^2 y/dx^2 -(m+n)dy/dx+mny

Answer 1

Given, $\bf{y=Ae^{mx}+Be^{nx}}$
we have to proof : $\bf{\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny}=0$

y = Ae^{mx} + Be^{nx} ---------6(1)
differentiate y with respect to x,
dy/dx = A. d(e^{mx})/dx + B. d(e^{nx})/dx
= A.m.e^{mx} + B.n.e^{nx} -----(2)

differentiate dy/dx with respect to x once again.
d²y/dx² = A.m. d(e^{mx})/dx + B.n. d(e^{nx})/dx
= A.m².e^{mx} + B.n².e^{nx} -------(3)

now, LHS = $\bf{\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny}$

from equations (2) and (3),

= (A.m².e^{mx} + B.n².e^{nx}) - (m + n)(A.m.e^{mx} + B.n.e^{nx}) + mny

= (A.m².e^{mx} + B.n².e^{nx}) - (A.m².e^{mx}+Amn.e^{mx} + B.n².e^{nx} + B.mn.e^{nx}) + mny

= -mn(A.e^{mx} + B.e^{nx}) + mny
= -mny + mny [ from equations (1),
= 0 = RHS $\textbf{\underline{hence proved}}$