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If $y=\mathrm{A} e^{m x}+\mathrm{Be}^{e^{m}}$, show that $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m r y=0$
If $y=\mathrm{A} e^{m x}+\mathrm{Be}^{e^{m}}$, show that $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m r y=0$
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.7 | Maths | NCERT
If $y=\mathrm{A} e^{m x}+\mathrm{Be}^{e^{m}}$, show that $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m r y=0$

Solution:

We need to Prove: $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$

$\frac{d y}{d x}=\mathrm{A} e^{m x} \frac{d}{d x}(m x)+\mathrm{B} e^{n x} \frac{d}{d x}(n x)\left[\because \frac{d}{d x} e^{f(x)}=e^{f(x)} \frac{d}{d x} f(x)\right]$

$\frac{d y}{d x}=\mathrm{A} m e^{m \mathrm{c}}+\mathrm{B} n e^{\pi x} \ldots(2)$

Find the derivate of equation (2)

$\frac{d^{2} y}{d x^{2}}=\mathrm{A} m e^{m \mathrm{c}} m+\mathrm{B} n e^{m x} \cdot n$

So now,

L.H.S. $=\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y$

(Using the eqs. (1), (2) and (3))

$=\mathrm{A} m^{2} e^{m x}+\mathrm{B} n^{2} e^{\prime x}-(m+n) \mathrm{A} m e^{\pi x}+\mathrm{B} n e^{m x}+m n\left(\mathrm{~A} e^{m x}+\mathrm{B} e^{m x}\right)$

$=A m^{2} e^{m x}+B n^{2} e^{m}-A m^{2} e^{m x}-B m n e^{m x}+A m n e^{m x}-B n^{2} e^{z x}+A m n e^{m x}+B m n e^{m}$

$=0$

$=$ R.H.S.

As a result hence proved.

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