Answer to Question #108726 in Differential Equations for sara alnaqbi

Question

y"−3y' + 2y = 3e−x−10cos3x, y(0) = 1, y'(0) = 2.

Answer 1

The general solution of the second order nonhomogeneous linear equation can be expressed in the form

$$y=y_c+Y$$

where $Y$ is any specific function that satisfies the nonhomogeneous equation, and $y_c=C_1y_1+C_2y_2$ is a general solution of the corresponding homogeneous equation

$$y''-3y'+2y=0$$

The Characteristic Equation

$$r^2-3r+2=0$$ $$(r-1)(r-2)=0$$

$r_1=1, r_2=2$

The general solution of the corresponding homogeneous equation is

$$y_c=C_1e^{x}+C_2e^{2x}$$

Use Method of Undetermined Coefficients

$$Y=Ae^{-x}+B\cos(3x)+C\sin(3x)$$

$$Y'=-Ae^{-x}-3B\sin(3x)+3C\cos(3x)$$

$$Y''=Ae^{-x}-9B\cos(3x)-9C\sin(3x)$$

$$Ae^{-x}-9B\cos(3x)-9C\sin(3x)+$$

$$+3Ae^{-x}+9B\sin(3x)-9C\cos(3x)+$$

$$+2Ae^{-x}+2B\cos(3x)+2C\sin(3x)=3e^{-x}-10\cos(3x)$$

$6A=3=>A=\dfrac{1}{2}$

$-7B-9C=-10$

$9B-7C=0$

$B=\dfrac{7}{13} , C=\dfrac{9}{13}$

$$Y=\dfrac{1}{2}e^{-x}+\dfrac{7}{13}\cos(3x)+\dfrac{9}{13}\sin(3x)$$

$$y=C_1e^{x}+C_2e^{2x}+\dfrac{1}{2}e^{-x}+\dfrac{7}{13}\cos(3x)+\dfrac{9}{13}\sin(3x)$$

$y(0)=1:$

$$1=C_1+C_2+\dfrac{1}{2}+\dfrac{7}{13}$$

$y'(0)=2:$

$$y'=C_1e^{x}+2C_2e^{2x}-\dfrac{1}{2}e^{-x}-\dfrac{21}{13}\sin(3x)+\dfrac{27}{13}\cos(3x)$$

$$2=C_1+2C_2-\dfrac{1}{2}+\dfrac{27}{13}$$

$$1=C_2-1+\dfrac{20}{13}$$

$$C_2=\dfrac{6}{13}$$

$$C_1=-\dfrac{1}{2}$$

$$y_p=-\dfrac{1}{2}e^{x}+\dfrac{6}{13}e^{2x}+\dfrac{1}{2}e^{-x}+\dfrac{7}{13}\cos(3x)+\dfrac{9}{13}\sin(3x)$$