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Find the general solution to the homogeneous differential equation frac{d^2y}{dt^2} - 8...

Question:

Find the general solution to the homogeneous differential equation

{eq}\frac{d^2y}{dt^2} - 8 \frac{dy}{dt} + 32y = 0 {/eq}

The solution has the form

{eq}y = C_1 f_1(t) + C_2f_2(t) {/eq} with

{eq}f_1(t) = ? {/eq}

and

{eq}f_2(t) = ? {/eq}

Complex Roots of Characteristic Equation

To find the solution of a homogeneous second order linear differential equation in the form {eq}Ay'' + By' + Cy = 0 {/eq}, where {eq}A, B \ and \ C {/eq} are constant real numbers, get the characteristic equation and solve for the roots of the characteristic equation. If the roots are complex numbers in the form {eq}r = \alpha \pm \beta i {/eq}, then the solution is in the form {eq}y = e^{\alpha x}(c_1\cos(\beta x) + c_2\sin (\beta x)) {/eq}, where {eq}c_1 \ and \ c_2 {/eq} are arbitrary constants.

Answer and Explanation:

The differential equation

{eq}\dfrac{d^2y}{dt^2} - 8 \dfrac{dy}{dt} + 32y = 0 {/eq}

can be written as

{eq}y'' - 8y' + 32y = 0 {/eq}

We need to find the characteristic equation of the differential equation by assuming the solution to be

{eq}y = e^{rt} {/eq}

The derivatives of this are

{eq}y' = re^{rt} \\ y'' = r^2e^{rt} {/eq}

Substitute these to the differential equation.

{eq}(r^2e^{rt}) - 8(re^{rt}) + 32(e^{rt}) = 0 \\ r^2e^{rt} - 8re^{rt} + 32e^{rt} = 0 {/eq}

Factor out and cancel {eq}e^{rt} {/eq} to get the characteristic equation.

{eq}e^{rt}(r^2 - 8r + 32) = 0 \\ r^2 - 8r + 32 = 0 {/eq}

To get the roots of the characteristic equation, we will use the quadratic formula.

{eq}r = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ r = \dfrac{-(-8)\pm \sqrt{(-8)^2-4(1)(32)}}{2(1)} \\ r = \dfrac{8\pm \sqrt{64 -128}}{2} \\ r = \dfrac{8\pm \sqrt{-64}}{2} \\ r = \dfrac{8\pm \sqrt{-1\cdot 64}}{2} \\ r = \dfrac{8\pm 8\sqrt{-1}}{2} \\ r = \dfrac{8\pm 8i}{2} \\ r = 4 \pm 4i {/eq}

Since the roots are complex number in the form {eq}r= \alpha \pm \beta i {/eq}, the form of the solution is

{eq}y = e^{\alpha t}(C_1\cos(\beta t) + C_2\sin (\beta t)) {/eq}

The solution is therefore:

{eq}y = e^{4t}(C_1\cos(4t) + C_2\sin (4t)) \\ y = C_1e^{4t}\cos(4t) + C_2e^{4t}\sin (4t) \\ {/eq}

The values of {eq}f_1(t) {/eq} and {eq}f_2(t) {/eq} are

{eq}\boldsymbol{f_1(t) = e^{4t}\cos(4t)} \\ \boldsymbol{f_2(t) = e^{4t}\sin(4t)} {/eq}


Learn more about this topic:

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Differential Calculus: Definition & Applications

from Calculus: Help and Review

Chapter 13 / Lesson 6
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