Question

Q. solve (D3−7D+12)y=x3−5x2

Answer 1

Answer on Question #64773 – Math – Differential Equations

Question

Solve $(D3 - 7D + 12)y = x3 - 5x2$

Solution

To solve a nonhomogeneous linear differential equation (DE)

$$(D^3 - 7D + 12)y = x^3 - 5x^2,$$

that is,

$$y''' - 7y' + 12y = x^3 - 5x^2$$

we must

1) find the complementary function $y_c$ that is the general solution of the associated homogeneous DE

$$y''' - 7y' + 12y = 0;$$

2) find any particular solution $y_p$ of the nonhomogeneous equation

$$y''' - 7y' + 12y = x^3 - 5x^2.$$

The general solution of the equation (1) is $y = y_c + y_p$.

1) To solve the associated homogeneous DE

$$y''' - 7y' + 12y = 0$$

we substitute into equation the solution as exponential function $y = e^{mx}$. We get

$$m^3 e^{mx} - 7m e^{mx} + 5e^{mx} = 0$$

or

$$(m^3 - 7m + 12)e^{mx} = 0.$$

Since $e^{mx}$ is never zero for real values of $x$, the last equation is satisfied only when $m$ is a solution or root of the third-degree polynomial equation

$$m^3 - 7m + 12 = 0.$$

This equation is called the auxiliary equation of the differential equation (2).

Analytical solution of this equation is a difficult task. Approximate numerical solution is $m_1 \approx 1.63 - 1.00i$, $m_2 \approx 1.63 + 1.00i$, $m_3 \approx -3.27$.

Thus we have three different roots of the auxiliary equation and the general solution of the associated homogeneous DE is

$$y_c = C_1 e^{(1.63 - 1.00i)x} + C_2 e^{(1.63 + 1.00i)x} + C_3 e^{-3.27x}.$$

Now we use Euler's formula $e^{i\beta} = \cos\beta + i\sin\beta$. Then we get

$$y_c = c_1 e^{1.63x} \cos x + c_2 e^{1.63x} \sin x + c_3 e^{-3.27x}$$

where $c_1, c_2, c_3$ are constants.

2) The particular solution of the nonhomogeneous equation will be found in the following form:

$$y_p = A x^3 + B x^2 + C x + D,$$

where $A, B, C, D$ are the undetermined coefficients.

Substituting this solution into the equation (1) and equating the coefficients of corresponding terms, we find $A, B, C, D$. Find derivatives $y'_p, y''_p, y'''_p$

$$y'_p = 3A x^2 + 2B x + C$$ $$y''_p = 6A x + 2B$$ $$y'''_p = 6A$$

Substituting the derivatives into the equation (1)

$$6A - 7(3Ax^2 + 2Bx + C) + 12(Ax^3 + Bx^2 + Cx + D) = x^3 - 5x^2$$

Equating the coefficients of corresponding terms, we find

$$\begin{cases} 12A = 1, \\ -21A + 12B = -5 \\ -14B + 12C = 0 \\ 6A - 7C + 12D = 0 \end{cases}$$ $$\begin{cases} A = \frac{1}{12} \approx 0.083, \\ -21 \cdot 0.083 + 12B = -5 \\ -14B + 12C = 0 \\ 6 \cdot 0.083 - 7C + 12D = 0 \end{cases}$$ $$\begin{cases} 12B = -5 + 1.743 = -3,257 \\ -14B + 12C = 0 \\ 0.5 - 7C + 12D = 0 \end{cases}$$ $$\begin{cases} B = -\frac{39}{144} \approx -0.271 \\ 14 \cdot 0.271 + 12C = 0 \\ \frac{1}{2} - 7C + 12D = 0 \end{cases}$$ $$\begin{cases} C \approx -0.316 \\ \frac{1}{2} + 7 \cdot 0.316 + 12D = 0 \end{cases}$$ $$D \approx -0.226$$

So we have $A \approx 0.083$, $B \approx -0.271$, $C \approx -0.316$, $D \approx -0.226$ and finally we get the general solution of differential equation (1):

$$y = c_1 e^{1.63x} \cos x + c_2 e^{1.63x} \sin x + c_3 e^{-3.27x} + 0.083x^3 - 0.271x^2 - 0.316x - 0.226.$$

Answer:

$$y = c_1 e^{1.63x} \cos x + c_2 e^{1.63x} \sin x + c_3 e^{-3.27x} + 0.083x^3 - 0.271x^2 - 0.316x - 0.226.$$

Answer provided by https://www.AssignmentExpert.com