Question

solve the second order differential equation

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

Answer 1

Answer to Question #49513 – Math – Differential Calculus | Equations

So, we have a next problem:

find the solution of an ordinary differential equation

$$y'' - 3y' + 2y = 3e^{-x} - 10\cos(3x),$$

that satisfies initial conditions (a Cauchy problem):

$$y(0) = 1, \quad y'(0) = 2$$

**Solution**

**The first step:** Let's find the particular solution $y_h$ of homogeneous equation:

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

Let's write the characteristic equation:

$$\lambda^2 - 3\lambda + 2 = 0.$$

Solving this equation we obtain two values of $\lambda$, namely $\lambda_1 = 1$ and $\lambda_2 = 2$.

So, $y_h = C_1 e^x + C_2 e^{2x}$, where $C_1$ and $C_2$ are unknown coefficients, which will be determined below.

**The second step:** Taking into account the right-hand side of equation (1), let's find the particular solution $y_p$ of the equation (1):

$$y_p = A\cos(3x) + B\sin(3x) + C e^{-x},$$

where $A$, $B$ and $C$ are unknown coefficients. Let's find these coefficients.

$$y_p' = -3A\sin(3x) + 3B\cos(3x) - C e^{-x}$$ $$y_p'' = -9A\cos(3x) - 9B\sin(3x) + C e^{-x}.$$

Substituting $y_p, y_p', y_p''$ for $y$ and its derivatives in the initial equation (1), we will obtain the next three equations to determine $A$, $B$ and $C$:

$$\begin{array}{l} -9A\cos(3x) - 9B\sin(3x) + C e^{-x} - 3(-3A\sin(3x) + 3B\cos(3x) - C e^{-x}) + 2(A\cos(3x) + B\sin(3x) + C e^{-x}) = \\ = 3e^{-x} - 10\cos(3x) \end{array}$$

Now let's collect the coefficients for $\cos(3x)$:

$$(-9A - 9B + 2A) = -10.$$

Similarly, collect the coefficients for $\sin(3x)$:

$$-9B + 9A + 2B = 0$$

In the same way, for $e^{-x}$:

$$C + 3 C + 2 C = 3 \Rightarrow C = \frac {1}{2}.$$

Solving the system of equations:

$$\left\{ \begin{array}{l} - 7 A - 9 B = - 1 0 \\ - 7 B + 9 A = 0 \end{array} \right.,$$

we will find $A = \frac{7}{13}, B = \frac{9}{13}$ .

Thus, $y_{p} = \frac{7}{13}\cos (3x) + \frac{9}{13}\sin (3x) + \frac{1}{2} e^{-x}$ .

The third step: Let's write the general solution of (1) as $y_{h} + y_{p}$ :

$$y = y _ {h} + y _ {p} = C _ {1} e ^ {x} + C _ {2} e ^ {2 x} + \frac {7}{1 3} \cos (3 x) + \frac {9}{1 3} \sin (3 x) + \frac {1}{2} e ^ {- x}.$$

The fourth step: Recall conditions (2) and solve the Cauchy problem, determine the coefficients $C_1$ and $C_2$ . From (3) we obtain

$$y ^ {\prime} = C _ {1} e ^ {x} + 2 C _ {2} e ^ {2 x} - \frac {2 1}{1 3} \sin (3 x) + \frac {2 7}{1 3} \cos (3 x) - \frac {1}{2} e ^ {- x}$$

Plug (3) and (4) into (2):

$$\left\{ \begin{array}{l} y (0) = 1 \\ y ^ {\prime} (0) = 2 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C _ {1} e ^ {0} + C _ {2} e ^ {0} + \frac {7}{1 3} + \frac {1}{2} e ^ {0} = 1 \\ C _ {1} e ^ {0} + 2 C _ {2} e ^ {0} + \frac {2 7}{1 3} - \frac {1}{2} e ^ {0} = 2 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C _ {1} + C _ {2} + \frac {7}{1 3} + \frac {1}{2} = 1 \\ C _ {1} + 2 C _ {2} + \frac {2 7}{1 3} - \frac {1}{2} = 2 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} C _ {1} = - \frac {1}{2} \\ C _ {2} = \frac {6}{1 3} \end{array} \right.$$

So, $y = -\frac{1}{2} e^{x} + \frac{6}{13} e^{2x} + \frac{7}{13}\cos (3x) + \frac{9}{13}\sin (3x) + \frac{1}{2} e^{-x}$ .

Answer: $y = -\frac{1}{2} e^{x} + \frac{6}{13} e^{2x} + \frac{7}{13}\cos (3x) + \frac{9}{13}\sin (3x) + \frac{1}{2} e^{-x}$ .

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