Find the solution of the differential equation y'' - 9y' + 18y = -3e^{-4t}

Question:

Find the solution of the differential equation {eq}y'' - 9y' + 18y = -3e^{-4t} {/eq}

Non Homogeneous Second Order Linear Differential Equation

A non homogeneous second order linear differential equation is a differential equation in the form {eq}Ay'' + By' + Cy = P(t) {/eq}, where {eq}A, B \ and \ C {/eq} are constant real numbers. This is solved by combining the complementary solution {eq}y_h {/eq}, or the solution to the homogeneous term {eq}Ay'' + By' + Cy = 0 {/eq} of the differential equation, and the particular solution {eq}y_p {/eq}, or the solution to the non homogeneous term {eq}P(t) {/eq} of the differential equation:

{eq}y(t) = y_h + y_p {/eq}

Answer and Explanation:

Complementary Solution

Let's begin by getting the solution to the corresponding homogeneous equation of the differential equation.

{eq}y'' - 9y' + 18y = 0 {/eq}

Assume the solution to be

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

The first and second derivatives of this are

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

Substitute these to the homogeneous differential equation.

{eq}(r^2e^{rt}) - 9(re^{rt}) + 18(e^{rt}) = 0 \\ r^2e^{rt} - 9re^{rt} + 18e^{rt} = 0 \\ e^{rt}(r^2 - 9r + 18) = 0 {/eq}

Divide both sides by {eq}e^{rt} {/eq} to get the characteristic equation.

{eq}r^2 - 9r + 18 = 0 {/eq}

The roots of this can be computed using the quadratic formula:

{eq}r = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ r = \dfrac{-(-9)\pm \sqrt{(-9)^2-4(1)(18)}}{2(1)} \\ r = \dfrac{9 \pm \sqrt{81-72}}{2} \\ r = \dfrac{9 \pm \sqrt{9}}{2} \\ r = \dfrac{9 \pm 3}{2} \\ r = \dfrac{9 + 3}{2}; \; r = \dfrac{9 - 3}{2} {/eq}

The values of the roots are

{eq}r = 6; \; r = 3 {/eq}

The complementary solution is therefore

{eq}y_h = c_1e^{6t} + c_2e^{3t} {/eq}

where {eq}c_1 \ and \ c_2 {/eq} are arbitrary coefficients.


Particular Solution

This is the solution to the non homogeneous term of the equation. Since the non homogeneous term is an exponential function, we can apply the method of undetermined coefficient. In this method, we will guess the differential equation based on the form of the non homogeneous term. We guess the particular solution to be

{eq}y_p = Ae^{-4t} {/eq}

We need to find the value of the unknown coefficient A. First, let's get the first and second derivative of this solution.

{eq}y_p' = -4Ae^{-4t} \\ y_p'' = 16Ae^{-4t} {/eq}

We will substitute these values to the differential equation.

{eq}y'' - 9y' + 18y = -3e^{-4t} \\ (16Ae^{-4t}) - 9(-4Ae^{-4t}) + 18(Ae^{-4t}) = -3e^{-4t} \\ 16Ae^{-4t} + 36Ae^{-4t} + 18Ae^{-4t} = -3e^{-4t} \\ 70Ae^{-4t} = -3e^{-4t} {/eq}

Equate the coefficients and solve for A.

{eq}70A = -3 \\ A = -\dfrac{3}{70} {/eq}

Therefore, the particular solution is

{eq}y_p = -\dfrac{3}{70}e^{-4t} {/eq}


General Solution

The general solution is solved by combining the complementary and particular solution.

{eq}y(t) = y_h + y_p \\ \boldsymbol{y(y) = c_1e^{6t} + c_2e^{3t} - \dfrac{3}{70}e^{-4t}} {/eq}


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