Question

Find the two values of k for which y(x)=e^{k x} ) is a solution of the differential equation \y^{\prime \prime}-11 y^{\prime}+30 y=0

smaller value = ____
larger value = _____

Answer 1

The two values of k for which y(x)=e^{kx} satisfies the differential equation y'' - 11y' + 30y = 0 are 5 and 6, with 5 being the smaller value and 6 the larger.

Step-by-step explanation:

To find the two values of k for which y(x)=e^{kx} is a solution of the differential equation y'' - 11y' + 30y = 0, we must substitute y(x) into the equation and solve for k.

Let's differentiate y(x) twice:

• First derivative: y' = ke^{kx}
• Second derivative: y'' = k^2e^{kx}

Substituting these into the differential equation gives:

k^2e^{kx} - 11ke^{kx} + 30e^{kx} = 0.

Factor out e^{kx}:

e^{kx}(k^2 - 11k + 30) = 0.

Since e^{kx} is never zero, we can divide by it:

k^2 - 11k + 30 = 0.

Factoring the quadratic equation:

(k-5)(k-6) = 0.

Therefore, the two values of k are 5 and 6, where 5 is the smaller value and 6 is the larger value.