Q14:

If $y = Ae^{mx} + Be^{nx}$, show that $\frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny = 0$

Step 1: Identifying the Primary Function and Differentiation

Given the function: $y = Ae^{mx} + Be^{nx}$

We perform the first differentiation with respect to $x$ using the Chain Rule, $\frac{d}{dx}[e^{kx}] = ke^{kx}$ [By the Exponential Differentiation Rule]:

$\frac{dy}{dx} = A \cdot \frac{d}{dx}(e^{mx}) + B \cdot \frac{d}{dx}(e^{nx})$

$\frac{dy}{dx} = Am e^{mx} + Bn e^{nx}$ --- (Equation 1)

Step 2: Performing the Second Differentiation

To obtain the second derivative, differentiate Equation 1 with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(Am e^{mx} + Bn e^{nx})$

$\frac{d^2y}{dx^2} = Am \cdot \frac{d}{dx}(e^{mx}) + Bn \cdot \frac{d}{dx}(e^{nx})$

$\frac{d^2y}{dx^2} = Am^2 e^{mx} + Bn^2 e^{nx}$ --- (Equation 2)

Step 3: Substituting into the Differential Equation

We must verify if the expression $\frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny$ equals zero. Substitute the expressions derived in Step 1 and Step 2:

LHS $= (Am^2 e^{mx} + Bn^2 e^{nx}) - (m + n)(Am e^{mx} + Bn e^{nx}) + mn(Ae^{mx} + Be^{nx})$

Step 4: Algebraic Expansion and Simplification

Expand the terms within the expression:

LHS $= Am^2 e^{mx} + Bn^2 e^{nx} - [m(Am e^{mx} + Bn e^{nx}) + n(Am e^{mx} + Bn e^{nx})] + mnAe^{mx} + mnBe^{nx}$

LHS $= Am^2 e^{mx} + Bn^2 e^{nx} - (Am^2 e^{mx} + Bmne^{nx} + Amn e^{mx} + Bn^2 e^{nx}) + Amne^{mx} + Bmne^{nx}$

Distribute the negative sign across the middle bracket:

LHS $= Am^2 e^{mx} + Bn^2 e^{nx} - Am^2 e^{mx} - Bmne^{nx} - Amn e^{mx} - Bn^2 e^{nx} + Amne^{mx} + Bmne^{nx}$

Step 5: Cancellation of Terms

Group terms by their exponential components to demonstrate the cancellation:

Every term cancels out perfectly [By the Commutative and Distributive Properties of Algebra].

LHS $= 0 = \text{RHS}$

Final Solution: The identity $\frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny = 0$ is proven for the given function.