Question

dx
If y = 3e23 + 2e3x, prove that
d?
dy
5
+ 6y = 0.
dx
dx2​

Answer 1

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Answer 2

Answer:

$y = 3e {}^{2x} + 2e {}^{3x}$

$\frac{dy}{dx} = 6e {}^{2x} + 6e {}^{3x}$

$\frac{d {}^{2} y}{dx {}^{2} } = 12e {}^{2x} + 18e {}^{3x}$

Condition:

$\frac{d {}^{2} y}{dx {}^{2} } - 5 \frac{dx}{dy} + 6y = 0$

$12e {}^{2x} + 18e {}^{3x} - 5(6e {}^{2x} + 6e {}^{3x} ) + 6(3e {}^{2x} + 2e {}^{3x} )$

$12e { }^{2x} + 18e {}^{3x} - 30e {}^{2x} - 30e {}^{3x} + 18e {}^{2x} + 12e {}^{3x}$

$30e {}^{2x} + 30e {}^{3x} - 30e {}^{2x} - 30e {}^{3x} = 0$

Hence 0 is the answer.

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