Question

Solve the DE (D^2-3D+2)y = 2 cos (2x+3)+2e^x​

Answer 1

Answer:

Solution: Given equation is. (D2 + 3D + 5) y = e2x. AE is D2 + 3D + 5 = 0. ∴. D = 3. 9 20. 3. 11. 2. 2. 2 i. - ±. -. - . = ±. ∴. CF = 3. 2. 1. 2. 11. 11 cos sin. 2. 2 x e c. x c x.

Answer 2

Answer:

$y =C_{1}e^{x}+C_{2}e^{2x} -\frac{cos(2x+3)}{10} -\frac{3sin(2x+3)}{10}-2xe^{x}.$

Step-by-step explanation:

Auxiliary equation:  

$\\m^{2} -3m +2 =0 \implies m=1,2.$

C.F. is given by,

$y_{cf} = c_{1}e^{x}+ c_{2}e^{2x} .$

Now we will solve for particular integral using known results,

$y_{pi} =\frac{1}{D^{2} -3D+2} (2cos(2x+3))+\frac{2}{(D-2)(D-1)}(e^{x}) ,\\y_{pi} =\frac{1}{D^{2} -3D+2} (e^{i(2x+3)}+e^{-i(2x+3)})+\frac{2}{(1-2)(D-1)}(e^{x}) ,\\y_{pi} =\frac{1}{(2i)^{2} -3(2i)+2} (e^{i(2x+3)})+\frac{1}{(-2i)^{2} -3(-2i)+2}( e^{-i(2x+3)})+\frac{2}{(1-2)(D-1)}(e^{x}) ,\\\\y_{pi} =\frac{e^{i(2x+3)}}{(2i)^{2} -3(2i)+2} +\frac{e^{-i(2x+3)}}{(-2i)^{2} -3(-2i)+2}+\frac{2}{(-1)(D-1)}(e^{x}) ,$

P.I. is given by,

$y_{pi} =\frac{e^{i(2x+3)}}{-2-6i} +\frac{e^{-i(2x+3)}}{-2+6i}-2xe^{x},\\\\y_{pi} =\frac{e^{i(2x+3)}}{-2-6i} +\frac{e^{-i(2x+3)}}{-2+6i}-2xe^{x},\\\\y_{pi} =-\frac{cos(2x+3)}{10} -\frac{3sin(2x+3)}{10}-2xe^{x},$

Finally, y = C.F + P.I

$y =C_{1}e^{x}+C_{2}e^{2x} -\frac{cos(2x+3)}{10} -\frac{3sin(2x+3)}{10}-2xe^{x}.$