Question

dx+y=cos3x
o yP(x)=1/10(−3sin3x−cos3x)
o yP(x)=1/10(−3sin3x+cos3x)
o yP(x)=1/10(3sin3x−cos3x)
o yP(x)=1/10(3sin3x+cos3x)

Answer 1

Answer on Question #79197 – Math – Differential Equations Question

$$\begin{array}{l} \frac{dy}{dx} + y = \cos 3x \\ y_p = \frac{1}{10}(-3 \sin 3x - \cos 3x) \\ y_p = \frac{1}{10}(-3 \sin 3x + \cos 3x) \\ y_p = \frac{1}{10}(3 \sin 3x - \cos 3x) \\ y_p = \frac{1}{10}(3 \sin 3x + \cos 3x) \\ \end{array}$$

Solution

First order non-homogeneous differential equation

$$\frac{dy}{dx} + y = \cos 3x$$

Solve the homogeneous equation

$$\begin{array}{l} \frac{d\bar{y}}{dx} + \bar{y} = 0 \\ \frac{d\bar{y}}{\bar{y}} = -dx \\ \int \frac{d\bar{y}}{\bar{y}} = -\int dx \\ \ln |\bar{y}| = -x + \ln C \\ \bar{y} = C e^{-x} \\ \text{Let } y_p = A \sin 3x + B \cos 3x \\ \end{array}$$

Then

$$\begin{array}{l} y_p' = 3A \cos 3x - 3B \sin 3x \\ y_p' + y_p = \cos 3x \\ 3A \cos 3x - 3B \sin 3x + A \sin 3x + B \cos 3x = \cos 3x \\ \left\{ \begin{array}{l} 3A + B = 1 \\ -3B + A = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 9A + B = 1 \\ A = 3B \end{array} \right. \Rightarrow \left\{ \begin{array}{l} A = \frac{3}{10} \\ B = \frac{1}{10} \end{array} \right. \\ y_p = \frac{1}{10}(3 \sin 3x + \cos 3x) \\ y = \bar{y} + y_p \\ y = C e^{-x} + \frac{1}{10}(3 \sin 3x + \cos 3x) \\ \end{array}$$

Answer: the fourth option is correct, $y_p = \frac{1}{10}(3 \sin 3x + \cos 3x)$,

$$y = C e^{-x} + \frac{1}{10}(3 \sin 3x + \cos 3x).$$

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