Question

Solve the following differential equation (3 marks)

d

2

y

dx2

− 6

dy

dx + 9y = x

2

e

3x

using the method of undetermined coefficients.

Answer 1

$$\dfrac{dy^2}{x^2}-6\dfrac{dy}{dx}+9y=x^2e^{3x}$$

Solve the correspondent homogeneous differential equation.

$$\dfrac{dy^2}{x^2}-6\dfrac{dy}{dx}+9y=0$$

The characteristic equation for this differential equation and its roots are

$$r^2-6r+9=0$$

$$r_{1,2}=3$$

The complementary solution is then

$$y_h(x)=C_1xe^{3x}+C_1e^{3x}$$

Find the particular solution of the non-homogeneous differential equation

$$y_p(x)=x^2(Ax^2+Bx+C)e^{3x}$$

Then

$$y_p'=e^{3x}(3Ax^4+3Bx^3+3Cx^2)$$

$$+e^{3x}(4Ax^3+3Bx^2+2Cx)$$

$$y_p''=e^{3x}(9Ax^4+9Bx^3+9Cx^2)$$

$$+e^{3x}(12Ax^3+9Bx^2+6Cx)$$

$$+e^{3x}(12Ax^3+9Bx^2+6Cx)$$

$$+e^{3x}(12Ax^2+6Bx+2C)$$

Substitute

$$e^{3x}(9Ax^4+9Bx^3+9Cx^2)$$

$$+e^{3x}(24Ax^3+18Bx^2+12Cx)$$

$$+e^{3x}(12Ax^2+6Bx+2C)$$

$$-e^{3x}(18Ax^4+18Bx^3+18Cx^2)$$

$$-e^{3x}(24Ax^3+18Bx^2+12Cx)$$

$$+e^{3x}(9Ax^4+9Bx^3+9Cx^2)$$

$$=x^2e^{3x}$$

$x^4: 9A-18A+9A=0$

$x^3: 9B+24A-18B-24A+9B=0$

$x^2: 9C+18B+12A-18C-18B+9C=1$

$=>A=\dfrac{1}{12}$

$x^1: 12C+6B-12C=0=>B=0$

$x^0: 2C=0=>C=0$

$$y_p=\dfrac{1}{12}x^4e^{3x}$$

The general solution of the given differential equation is

$$y(x)=C_1xe^{3x}+C_1e^{3x}+\dfrac{1}{12}x^4e^{3x}$$