Question

5.
6. Find the differential
(i) y = a e ^3x + be^x​

Answer 1

Answer:

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Answer 2

Answer:

correct answer :

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Step-by-step explanation:

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$y = a {e}^{3x} + b {e}^{x}$

$by \: getting \: both \: \: log_{e} \: \: sides$

$log_{e}y = log_{e}a {e}^{3x} + log_{e}b {e}^{x}$

$log_{e}y = 3x log_{e}ae + x log_{e}be$

By differentiate both sides dy/dx

$\frac{1}{y} \frac{dy}{dx} = 3x \frac{dy}{dx} logae + logea \frac{dy}{dx}3x + x \: \frac{dy}{dx}logbe + logbe \frac{dy}{dx}x$

$\frac{1}{y} \times \frac{dy}{dx} = 3x \times a + x \times b ............... { log_{e} }^{e} = 1$

$\frac{dy}{dx} = y(3xa + xb)$

$\frac{dy}{dx} = xy(3a + b)$

$put \: the \: y \: value \: in \: this \: equation \: because \: it \: is \: implicts \: function$

Required answer ::

$\frac{dy}{dx} = {ae}^{x} + b {e}^{x} (3xa + bx)$

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