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alekssr [168]
3 years ago
9

Verify that the indicated family of functions is a solution of the given differential equation. Assume an appropriate interval I

of definition for each solution.
d^2y/ dx^2 − 6 dy/dx + 9y = 0; y = c1e3x + c2xe3x When y = c1e3x + c2xe3x,
Mathematics
1 answer:
Rufina [12.5K]3 years ago
3 0

<em>y''</em> - 6<em>y'</em> + 9<em>y</em> = 0

If <em>y</em> = <em>C₁ </em>exp(3<em>x</em>) + <em>C₂</em> <em>x</em> exp(3<em>x</em>), then

<em>y'</em> = 3<em>C₁ </em>exp(3<em>x</em>) + <em>C₂</em> (exp(3<em>x</em>) + 3<em>x</em> exp(3<em>x</em>))

<em>y''</em> = 9<em>C₁ </em>exp(3<em>x</em>) + <em>C₂</em> (6 exp(3<em>x</em>) + 9<em>x</em> exp(3<em>x</em>))

Substituting these into the DE gives

(9<em>C₁ </em>exp(3<em>x</em>) + <em>C₂</em> (6 exp(3<em>x</em>) + 9<em>x</em> exp(3<em>x</em>)))

… … … - 6 (3<em>C₁ </em>exp(3<em>x</em>) + <em>C₂</em> (exp(3<em>x</em>) + 3<em>x</em> exp(3<em>x</em>)))

… … … + 9 (<em>C₁ </em>exp(3<em>x</em>) + <em>C₂</em> <em>x</em> exp(3<em>x</em>))

= 9<em>C₁ </em>exp(3<em>x</em>) + 6<em>C₂ </em>exp(3<em>x</em>) + 9<em>C₂ x</em> exp(3<em>x</em>))

… … … - 18<em>C₁ </em>exp(3<em>x</em>) - 6<em>C₂</em> (exp(3<em>x</em>) - 18<em>x</em> exp(3<em>x</em>))

… … … + 9<em>C₁ </em>exp(3<em>x</em>) + 9<em>C₂</em> <em>x</em> exp(3<em>x</em>)

= 0

so the provided solution does satisfy the DE.

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The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.

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