Answer to Question #222721 in Differential Equations for frank

Question

y''-4y'+3y=9x³+4 y(0)=6 y'(0)=8

Answer 1

Given Differential equation y'' - 4y'+3y = 9x³ + 4 with y(0) = 6, y'(0) = 8

Solution:

Auxiliary equation corresponding to this given Differential equation is r²- 4r+3=0

$\implies$ (r-3)(r-1) =0

$\implies$ r=3 and r=1

therefore, y_h = c₁e^3x + c₂e^x

Now to find particular integral (particular solution) = y_p

Let y_p = Ax³+Bx²+Cx+D

$\implies$ y_p' = 3Ax²+2Bx+C

$\implies$ y_p'' = 6Ax+2B

plugging this into given differential equation, y_p'' - 4y_p'+3y_p = 9x³ + 4

$\implies$ (6Ax+2B) - 4( 3Ax²+2Bx+C) + 3(Ax³+Bx²+Cx+D) = 9x³ + 4

$\implies$ 3Ax³- 12Ax²+3Bx²+6Ax-8Bx+3Cx+2B-4C+3D = 9x³ + 4

comparing both sides, we get A = 3, B = 12,C = 26, D = 28

hence y_p = 3x³ + 12x² + 26x + 28

the general solution of the given differential equation is y = y_h + y_p

y = c₁e^3x + c₂e^x + 3x³ + 12x² + 26x + 28

y' = 3c₁e^3x + c₂e^x + 9x2 + 24x + 26

Also given y(0) = 6, y'(0) = 8, plugging these conditions we get

6 = c₁ + c₂ + 28 $\implies$ c₁ + c₂ = -22

and 8 = 3c₁ + c₂ + 26 $\implies$ 3c₁ + c₂ = -18

By solving we get, c₁ = 2 & c₂ = -24

The solution of the differential equation is y = 2e^3x - 24e^x + 3x³ + 12x² + 26x + 28