Question

(d2-d)y=12ex+8sinx-2x​

Answer 1

Step-by-step explanation:

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Answer 2

The solution to the given differential equation is $y = A + B e^x + 12xe^x - 4 sinx + 4 cosx + x^{2} + 2x$

Step-by-step explanation:

Given: Differential equation $(D^{2} - D)y=12e^x+8sinx-2x$

To Find: solution of the given differential equation

Solution:

• Solution of the differential equation $(D^{2} - D)y=12e^x+8sinx-2x$

The complementary factor can be determined as,

$\Rightarrow m^{2} - m = 0 \Rightarrow m (m - 1) = 0 \Rightarrow m = 0, 1$

$\Rightarrow y_{_{CF}} = A + B e^x$

And, the particular integral is,

$\Rightarrow y_{_{PI}} = \dfrac{1}{D^2 - D} (12e^x + 8 sinx - 2x)$

$\Rightarrow y_{_{PI}} = 12 \dfrac{1}{D^2 - D} e^x + 8 \dfrac{1}{D^2 - D} sinx - 2\dfrac{1}{D^2 - D} x$

Now, using the rule $\frac{1}{f(D)}e^{ax} = x \frac{1}{f'(a)}e^{ax} \text{ for } f(a) = 0$ in the first term, $\frac{1}{f(D^2)}sinax = \frac{1}{f(-a^2)}sin{ax}$ in the second term, and $\frac{1}{f(D)}x^n = [f(D)]^{-1} x^n$ in the third term of the above expression, we will get;

$\Rightarrow y_{_{PI}} = 12 x e^x + 8 \dfrac{1}{(-1 - D)} sinx + \dfrac{2}{D} [1-D]^{-1} x$

$\Rightarrow y_{_{PI}} = 12 x e^x - 8 \dfrac{1 - D}{(1 - D^2)} sinx + \dfrac{2}{D} [1 + D + D^{2}] x$

$\Rightarrow y_{_{PI}} = 12 x e^x - 8 \dfrac{1 - D}{(1 - (-1))} sinx + \dfrac{2}{D} [x + Dx + D^{2}x]$

$\Rightarrow y_{_{PI}} = 12 x e^x - 4 [sinx - D(sinx)] + \dfrac{2}{D} [x + 1 + 0]$

$\Rightarrow y_{_{PI}} = 12 x e^x - 4 [sinx - cosx] + 2 \Big[ \dfrac{x^{2} }{2} + x \Big]$

$\Rightarrow y_{_{PI}} = 12 x e^x - 4sinx + 4cosx + x^{2} + 2x$

Hence, the solution to the given differential equation is $y = y_{_{CF}} + y_{_{PI}}$$\Rightarrow y = A + B e^x + 12xe^x - 4 sinx + 4 cosx + x^{2} + 2x$