Question

d^3y/dx^3-3 d^2y/dx^3+ 4 dy/dx -2y=e^x+cosx

Answer 1

Solution.

$$\frac{d^3y}{dx^3}-3\frac{d^2y}{dx^2}+4\frac{dy}{dx}-2y=e^x+\cos⁡{x},$$

or

$$y'''-3y''+4y'-2y=e^x+\cos{x}.$$

This is a linear differential equation of the third-order with constant coefficients. The general solution of this equation is found as the sum of the general solution $\tilde{y}$ of the corresponding homogeneous equation and some particular integral solution $y_*$ of the inhomogeneous equation: $y=\tilde{y}+y_*.$

Сompose and solve the characteristic equation:

$$\lambda^3-3\lambda^2+4\lambda-2=0, \newline(\lambda-1)(\lambda^2-2\lambda+2)=0, \newline \lambda_1=1, \text{or } \lambda^2-2\lambda+2=0, \newline \text{ } \lambda_2=1+i, \text{ } \lambda_3=1-i.$$

Therefore, $\tilde{y}=C_1e^{\lambda_1x}+C_2e^{ax}\cos{bx}+C_3e^{ax}\sin{bx},$

where $a$ is a real part, and $b$ is an imaginary part of complex numbers $\lambda_2 \text{ and } \lambda_3.$ So,

$$\newline \tilde{y}=C_1e^x+C_2e^x\cos{x}+C_3e^x\sin{x}.$$

The particular integral solution $y_*$ is writte in the form:

$$y_∗=y_1+y_2​.$$

Since the right-hand side contains a $\cos{x}$ , we look for a particular integral solution $y_1$ ​in the form:

$$y_1=A\cos{x}+B\sin{x}.$$

Using the method of undetermined coefficients we find that $A=\frac{1}{10}$ and $B=\frac{3}{10}.$ So, $y_1=\frac{1}{10}\cos{x}+\frac{3}{10}\sin{x}.$

Since the right-hand side also contains a $e^{x}$ and 1 is characteristic number, we find a particular integral solution $y_2$ ​in the form:

$$y_2=Cxe^{x}.$$

Using the method of undetermined coefficients we find that $C=1.$

Therefore, $y_2=xe^{x}.$

As follows, the particular integral solution of $\frac{d^3y}{dx^3}-3\frac{d^2y}{dx^2}+4\frac{dy}{dx}-2y=e^x+\cos⁡{x},$ is

$$y_*=\frac{1}{10}\cos{x}+\frac{3}{10}\sin{x}+xe^{x},$$

and the general solution of the same equation is

$$y=C_1e^x+C_2e^x\cos{x}+C_3e^x\sin{x}+xe^x+\frac{1}{10}\cos{x}+\frac{3}{10}\sin{x}.$$

Answer.

$y=C_1e^x+C_2e^x\cos{x}+C_3e^x\sin{x}+xe^x+\frac{1}{10}\cos{x}+\frac{3}{10}\sin{x}.$