Question

Y"+4y'-y=14+6e2x

Answer 1

The homogeneous differential equation

$$y''+4y'-y=0$$

The corresponding (auxiliary) equation

$$r^2+4r-1=0$$

$$D=(4)^2-4(1)(-1)=20$$

$$r=\dfrac{-4\pm\sqrt{20}}{2(1)}=-2\pm \sqrt{5}$$

The general solution of the homogeneous differential equation is

$$y=C_1e^{(-2-\sqrt{5})x}+C_2e^{(-2+\sqrt{5})x}$$

Find the particular solution of the non-homogeneous differential equation

$$y_p=A+Be^{2x}$$

$$y_p'=2Be^{2x}$$

$$y_p''=4Be^{2x}$$

Substitute

$$4Be^{2x}+4(2Be^{2x})-(A+Be^{2x})=14+6e^{2x}$$

$$-A+11Be^{2x}=14+6e^{2x}$$

$$A=-14, B=\dfrac{6}{11}$$

Then

$$y_p=-14+\dfrac{6}{11}e^{2x}$$

The general solution of the given non-homogeneous differential equation is

$$y=y_h+y_p$$

$$y=C_1e^{(-2-\sqrt{5})x}+C_2e^{(-2+\sqrt{5})x}-14+\dfrac{6}{11}e^{2x}$$