Question

(D^4+2D^3-3D^2)Y=3e^2x+4sinx.

Answer 1

Answer on Question # 82704 - Math - Differential Equations

Question 1.

$$y ^ {(4)} + 2 y ^ {(3)} - 3 y ^ {\prime \prime} = 3 e ^ {2 x} + 4 \sin x$$

Answer.

$$y (x) = c _ {1} e ^ {- 3 x} + c _ {2} e ^ {x} + c _ {3} x + c _ {4} + \frac {3}{2 0} e ^ {2 x} + \frac {4}{5} \sin x + \frac {2}{5} \cos x,$$

where $c_{1},c_{2},c_{3},c_{4}\in \mathbb{R}$.

Problem solving steps. 1) First, solve the corresponding homogeneous equation

$$y ^ {(4)} + 2 y ^ {(3)} - 3 y ^ {\prime \prime} = 0:$$

characteristic equation $\lambda^4 + 2\lambda^3 - 3\lambda^2 = 0$, $\lambda_1 = -3$, $\lambda_2 = 1$, $\lambda_3 = \lambda_4 = 0$;

then solutions are $y_0(x) = c_1e^{-3x} + c_2e^x + c_3x + c_4$, where $c_1, c_2, c_3, c_4 \in \mathbb{R}$.

2) Next, set up a trial function (partial solution) $y_{p}(x) = ae^{2x} + b\sin x + c\cos x$ by copying the structure of RHS of the nonhomogeneous equation and substitute into the nonhomogeneous equation:

$$2 0 a e ^ {2 x} + (4 b + 2 c) \sin x + (4 c - 2 b) \cos x = 3 e ^ {2 x} + 4 \sin x,$$ $$a = \frac {3}{2 0}, b = \frac {4}{5}, c = \frac {2}{5},$$ $$y _ {p} (x) = \frac {3}{2 0} e ^ {2 x} + \frac {4}{5} \sin x + \frac {2}{5} \cos x.$$

3) The general solutions to the nonhomogeneous equation are $y(x) = y_0(x) + y_p(x)$.

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