Question

Q5. If y = aemx + be-mx, Prove that d2y/dx2 = m2y.

Answer 1

$y = a {e}^{mx} + b {e}^{ - mx} \\ \frac{dy}{dx} = am {e}^{mx} - bm {e}^{ - mx} \\ \frac{ {d}^{2}y }{d {x}^{2} } = a {m}^{2} {e}^{mx} + b {m}^{2} {e}^{ - mx} \\ = {m}^{2} (a {e}^{mx} + b{e}^{ - mx} ) \\ \frac{ {d}^{2} y}{d {x}^{2} } = {m}^{2} y$

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Answer 2

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$= > y = a {e}^{mx} + b {e}^{ - mx} \\ \\ differentiating \: \: both \: \: sides \: \: by \: x \\ \\ = > \frac{dy}{dx} = \frac{d(a {e}^{mx} )}{dx} + \frac{d(b {e}^{ - mx}) }{dx} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = a \frac{d( {e}^{mx}) }{dx} + b \frac{d( {e}^{ - mx} )}{dx} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = am {e}^{mx} - bm {e}^{ - mx} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = m( a{e}^{mx} - b {e}^{ - mx} ) \\ \\ again \: \: differentiating \: \: both \: \: sides \: by \: x \\ \\ = > \frac{ {d}^{2}y }{d {x}^{2} } = m( \frac{d(a {e}^{mx} )}{dx} - \frac{d(b {e}^{ - mx} )}{dx} ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = m(am {e}^{mx} - ( - bm {e}^{ - mx} )) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {m}^{2} (a {e}^{mx } + b {e}^{ - mx} ) = {m}^{2} y$
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