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Find the two values of k for which y(x) = e^kx is a solution of the differential equation y'' - 20y' + 91y = 0. Preview smaller value = larger value = Preview

User Mabead
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1 Answer

4 votes

Answer:

The values of k are

1) k = 7.

2) k= 13

Explanation:

The given differential equation is


y''-20y'+91y=0

Now since it is given that
y=e^(kx) is a solution thus it must satisfy the given differential equation thus we have


(d^2)/(dx^2)(e^(kx))-20(d)/(dx)e^(kx)+91e^(kx)=0\\\\k^(2)\cdot e^(kx)-20\cdot k\cdot e^(kx)+91e^(kx)=0\\\\e^(kx)(k^(2)-20k+91)=0\\\\k^(2)-20k+91=0

This is a quadratic equation in 'k' thus solving it for k we get


k=(20\pm √((-20)^2-4\cdot 1\cdot 91))/(2)\\\\\therefore k=7,k=13

User Sloloem
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