Question

dx
dx²
Find the differential equation which has y=a cos (mx+b) for its complete integral, a and b being
arbitrary constants.
v=Ae2+ Bex
differ​

Answer 1

Answer:

Question:-

• Find the differential equation which has y=a cos (mx+b), a and b being arbitrary constants.

$\huge{\boxed{\rm{Answer}}}$

$\large{\boxed{\boxed{\sf{Given \: that}}}}$

• y=a cos (mx+b), a and b being arbitrary constants.

$\large{\boxed{\boxed{\sf{To \: find}}}}$

• The differential equation

$\large{\boxed{\boxed{\sf{Solution}}}}$

$\bf \:y=a \: cos (mx+b)........(1)$

Differentiate w. r. t. x

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{d}{dx}a cos (mx+b)$

$\bf\implies \:\dfrac{dy}{dx} = - am \: sin(mx + b)$

Differentiate w.r.t. x

$\bf\implies \:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - am \times mcos(mx + b)$

$\bf\implies \:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - {m}^{2} \times \: a cos (mx+b),$

$\bf\implies \:\dfrac{ {d}^{2}y }{ {dx}^{2} } = - {m}^{2} y$

$\bf\implies \:\dfrac{ {d}^{2}y }{ {dx}^{2} } + {m}^{2} y=0$

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