Question

Consider the differential equation, a2y''(x) + a1y'(x) + a0y(x) = 0, with coefficients a2 = 16, a1 = 56, a0 = 149. (a) What is the basis for the general solution of the differential equation? Enter the functions separated by semi-colons, with no extra spaces, for example, exp(2x);exp(-2x/3). Basis: exp(-7/4x)cos(5/2x);exp(- (b) Enter the general solution of the differential equation. You may use c1 and c2, or, k1 and k2 as your scalars. y(x) = e^(-7/4x)(c1cos(5/2x) + c2sin(5/2x))

          Consider the differential equation, a2y''(x) + a1y'(x) + a0y(x) = 0, with coefficients
a2 = 16, a1 = 56, a0 = 149.
(a) What is the basis for the general solution of the differential equation? Enter the functions separated by semi-colons, with no extra spaces, for example, exp(2*x);exp(-2*x/3).
Basis: exp(-7/4x)cos(5/2x);exp(-
(b) Enter the general solution of the differential equation. You may use c1 and c2, or, k1 and k2 as your scalars.
y(x) = e^(-7/4x)(c1cos(5/2x) + c2sin(5/2x))

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Consider the differential equation, a2y”(x) + a1y'(x) + a0y(x) = 0, with coefficients
a2 = 16, a1 = 56, a0 = 149.
(a) What is the basis for the general solution of the differential equation? Enter the functions separated by semi-colons, with no extra spaces, for example, exp(2*x);exp(-2*x/3).
Basis: exp(-7/4x)cos(5/2x);exp(-
(b) Enter the general solution of the differential equation. You may use c1 and c2, or, k1 and k2 as your scalars.
y(x) = e^(-7/4x)(c1cos(5/2x) + c2sin(5/2x))

Added by Juan Jos- U.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert James Kiss

Step 1

Now, we need to find the characteristic equation: $$16r^2 + 56r + 149 = 0$$ We can solve this quadratic equation for $r$ using the quadratic formula: $$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Plugging in the coefficients, we get: $$r = \frac{-56 \pm Show more…

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Consider the differential equation, a2y''(x) + a1y'(x) + a0y(x) = 0, with coefficients a2 = 16, a1 = 56, a0 = 149. (a) What is the basis for the general solution of the differential equation? Enter the functions separated by semi-colons, with no extra spaces, for example, exp(2*x);exp(-2*x/3). Basis: exp(-7/4x)cos(5/2x);exp(-7/4x)sin(5/2x) (b) Enter the general solution of the differential equation. You may use c1 and c2, or, k1 and k2 as your scalars. y(x) = e^(-7/4x)(c1cos(5/2x) + c2sin(5/2x))

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