Applications of Derivatives Question - None

Question

If $y = Ae^{mx} + Be^{nx}$, then the value of $\frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny$ is equal to

Accepted Answer

Solution:

Given, $y = Ae^{mx} + Be^{nx}$

Differentiating w.r.t. $x$:

$\frac{dy}{dx} = Ame^{mx} + Bne^{nx}$

Differentiating again w.r.t. $x$:

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

Now substituting in the expression:

$\frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny$

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

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

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

$= Ae^{mx}(0) + Be^{nx}(0) = 0$

∴ Answer is (b) 0

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