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ICE Princess25 [194]
3 years ago
7

Solve the differential equation

0y%5Csqrt%7B3x%5E%7B2%7D%20-%201%20%7D" id="TexFormula1" title="\frac{1}{6x} * \frac{dy}{dx} = y\sqrt{3x^{2} - 1 }" alt="\frac{1}{6x} * \frac{dy}{dx} = y\sqrt{3x^{2} - 1 }" align="absmiddle" class="latex-formula">
A. e^{2/3} \sqrt{(3x^{2} -1)^{3}}
B. C*e^{2/3} \sqrt{(3x^{2} -1)^{3}}
C. None of these
Mathematics
1 answer:
ExtremeBDS [4]3 years ago
6 0

This ODE is separable as

1/(6<em>x</em>) d<em>y</em>/d<em>x</em> = <em>y</em> √(3<em>x</em> ² - 1)

→   d<em>y </em>/ <em>y</em> = 6<em>x</em> √(3<em>x</em> ² - 1) d<em>x</em>

Integrate both sides:

∫ d<em>y </em>/ <em>y</em> = ∫ 6<em>x</em> √(3<em>x</em> ² - 1) d<em>x</em>

<em />

The left side is trivial. For the right side, substitute <em>u</em> = 3<em>x</em> ² - 1 and d<em>u</em> = 6<em>x</em> d<em>x</em> :

∫ d<em>y </em>/ <em>y</em> = ∫ √<em>u</em> d<em>u</em>

ln|<em>y </em>| = <em>u</em> ³′² + <em>C</em>

(that is, <em>u</em> is raised to the 3/2 power)

ln|<em>y</em> | = (3<em>x</em> ² - 1)³′² + <em>C</em>

<em />

Solve for <em>y</em> by taking the exponential of both sides:

exp(ln|<em>y</em> |) = exp((3<em>x</em> ² - 1)³′² + <em>C</em> )

<em>y</em> = exp((3<em>x</em> ² - 1)³′²) × exp(<em>C</em> )

<em>y</em> = <em>C</em> exp((3<em>x</em> ² - 1)³′²)

which can be written as

<em>y</em> = <em>C</em> exp(√((3<em>x</em> ² - 1)³))

which makes the answer none of these; this solution can't be expressed as either option given in A or B.

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