What is the solution of D.E (D2 – 2D)y = ex sin⁡x when solved using the method of undetermined coefficients?

Question

What is the solution of D.E (D2 – 2D)y = ex  sin⁡x when solved using the method of undetermined coefficients?

TuteeHUB · Accepted Answer

$y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x  cos x}{2}$

TuteeHUB · Answer

$ y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x (xsin x + cos⁡x)}{2}$

TuteeHUB · Answer

$y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x sin x}{2}$

TuteeHUB · Answer

$y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x  cos x}{2}$

TuteeHUB · Answer

$y =   c_1 +  c_2 \,e^{2x} –   \frac{e^x  (sin x + x cos⁡x)}{4}$

1.	What is the solution of D.E (D2 – 2D)y = ex sin⁡x when solved using the method of undetermined coefficients?
A.	\( y = c_1 + c_2 \,e^{2x} – \frac{e^x (xsin x + cos⁡x)}{2}\)
B.	\(y = c_1 + c_2 \,e^{2x} – \frac{e^x sin x}{2}\)
C.	\(y = c_1 + c_2 \,e^{2x} – \frac{e^x cos x}{2}\)
D.	\(y = c_1 + c_2 \,e^{2x} – \frac{e^x (sin x + x cos⁡x)}{4}\)
Answer» C. \(y = c_1 + c_2 \,e^{2x} – \frac{e^x cos x}{2}\)

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