Answer to Question #162123 in Differential Equations for Adil Rajpoot

Question

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Find the particular integral solution of  d^3y/dx^3 + 4(dy/dx)=x+3cosx+ e^-2x

Answer 1

Solution.

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

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:

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

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

where $a$ - real part, and $b$ - imaginary part of a complex number $\lambda_2 \text{ and } \lambda_3.$ So,

$$\newline \tilde{y}=C_1+C_2\cos{2x}+C_3\sin{2x}.$$

The particular integral solution $y_*$ write in the form:

$$y_∗=y_1+y_2+y_3​.$$

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 indefinite coefficients we find that $A=0$ and $B=1.$ So, $y_1=\sin{x}.$

Since the right-hand side also contains a $e^{-2x}$ , we find a particular integral solution $y_2$ ​in the form:

$$y_2=Ce^{-2x}.$$

Using the method of indefinite coefficients we find that $C=-\frac{1}{16}.$

Therefore, $y_2=-\frac{1}{16}e^{-2x}.$

The number is characteristic, so $y_3$ is written as

$$y_3=x(Dx+E).$$

Using the method of indefinite coefficients we find that $D=\frac{1}{8}$ and $E=0.$ So, $y_3=x(\frac{1}{8}x+0)=\frac{1}{8}x^2.$

As follows, the particular integral solution of $\frac{d^3y}{dx3}+4\frac{dy}{dx}=x+3\cos⁡{x}+e^{−2x}$ is

$$y_*=\sin{x}-\frac{1}{16}e^{-2x}+\frac{1}{8}x^2,$$

and the general solution of the same equation is

$$y=C_1+C_2\cos{2x}+C_3\sin{2x}+\sin{x}-\frac{1}{16}e^{-2x}+\frac{1}{8}x^2.$$

Answer.

$y_*=\sin{x}-\frac{1}{16}e^{-2x}+\frac{1}{8}x^2,$ where $y_*$ is the particular integral solution.