Question

Find the general solution of the following differential equation
y''-4y'+3y=2*exp(x)*(1+2x)+x

Answer 1

Answer on Question #58663 – Math – Differential Equations

Question

Find the general solution of the following differential equation

$$y'' - 4y' + 3y = 2(1 + 2x)e^x + x$$

Solution

The auxiliary equation is

$$\lambda^2 - 4\lambda + 3 = 0.$$

Its solutions are

$$\lambda_1 = \frac{4 - \sqrt{16 - 12}}{2} = 1, \quad \lambda_2 = \frac{4 + 2}{2} = 3.$$

The general solution of the homogeneous differential equation $y'' - 4y' + 3y = 0$ is

$$Y = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}$$ $$Y = c_1 e^x + c_2 e^{3x},$$

where $c_1, c_2$ are arbitrary real constants.

The general solution of the non-homogeneous differential equation

$$y'' - 4y' + 3y = 2(1 + 2x)e^x + x$$

is

$$y = Y + \bar{y},$$

$Y$ is the general solution of the homogeneous differential equation;

$\bar{y}$ is a particular solution of the non-homogeneous differential equation.

We use the method of undetermined coefficients.

Because $e^x = e^{x\lambda_1}$, we search a particular solution in the following form:

$$\bar{y} = (Ax^2 + Bx)e^x + Cx + D = Ax^2 e^x + Bx e^x + Cx + D$$ $$\tilde{y'} = (2Ax + B)e^x + (Ax^2 + Bx)e^x + C = Ax^2 e^x + (2A + B)x e^x + B e^x + C$$ $$\widetilde{y''} = 2Axe^x + Ax^2e^x + (2A + B)e^x + (2A + B)xe^x + Be^x = Ax^2e^x + 2(A + B)e^x + (4A + B)xe^x$$

Plug $\tilde{y}, \widetilde{y'}, \widetilde{y''}$ into the initial non-homogeneous differential equation:

$$\begin{aligned} Ax^2e^x + 2(A + B)e^x + (4A + B)xe^x - 4(Ax^2e^x + (2A + B)xe^x + Be^x + C) + \\ + 3(Ax^2e^x + Bxe^x + Cx + D) = 2(1 + 2x)e^x + x \\ - 4Axe^x + 2(A + B)e^x + 3Cx + 3D - 4C = 4xe^x + 2e^x + x \end{aligned}$$

Equate like terms and get the following system of equations:

$$\left\{ \begin{array}{l} -4A = 4 \\ 2(A - B) = 2 \\ 3C = 1 \\ 3D - 4C = 0 \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} A = -1 \\ B = -2 \\ C = \frac{1}{3} \\ D = \frac{4}{9} \end{array} \right.$$

Then

$$\tilde{y} = -x^2e^x - 2xe^x + \frac{x}{3} + \frac{4}{9}.$$

Finally get

$$y = Y + \tilde{y} = c_1e^x + c_2e^{3x} - x^2e^x - 2xe^x + \frac{x}{3} + \frac{4}{9}.$$

Answer: $y = c_1e^x + c_2e^{3x} - x^2e^x - 2xe^x + \frac{x}{3} + \frac{4}{9}$.

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