Question

Find the particular integral of dy/dx+y=cos3x

Answer 1

Answer on Question #69644 - Math - Differential Equations

Find the particular integral of $\frac{dy}{dx} + y = \cos 3x$ .

Solution

The corresponding homogenous equation is $\frac{dy}{dx} + y = 0$ . Its characteristic equation is $\mu + 1 = 0$ with the root $\mu = -1$ . With respect to it and considering the right part of the equation is $\cos 3x$ , the particular integral is

$$y _ {p} (x) = \alpha \cos 3 x + \beta \sin 3 x,$$

where $\alpha$ and $\beta$ are the coefficients to be evaluated.

The derivative of the particular integral is

$$\frac {d y _ {p}}{d x} = - 3 \alpha \sin 3 x + 3 \beta \cos 3 x,$$

Substituting the particular integral and its derivative into the equation we get

$$\begin{array}{l} \frac {d y _ {p}}{d x} + y _ {p} = (- 3 \alpha \sin 3 x + 3 \beta \cos 3 x) + (\alpha \cos 3 x + \beta \sin 3 x) = \\ = (\alpha + 3 \beta) \cos 3 x + (\beta - 3 \alpha) \sin 3 x = \cos 3 x. \\ \end{array}$$

To evaluate the unknown coefficients we get the system

$$\begin{array}{l} \alpha + 3 \beta = 1 \\ \beta - 3 \alpha = 0. \\ \end{array}$$

From the second equation we have $\beta = 3\alpha$ , thereby from the first equation we have $\alpha + 9\alpha = 1$ . The solution is $\alpha = \frac{1}{10}$ , thereby $\beta = 3 \cdot \frac{1}{10} = \frac{3}{10}$ .

The particular integral is $y_{p}(x) = \frac{1}{10}\cos 3x + \frac{3}{10}\sin 3x$ .

Answer: $y_{p}(x) = \frac{1}{10}\cos 3x + \frac{3}{10}\sin 3x$ .

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