Question

find the general solution of the following y11 -5y1+6y=2ex+6x-5​

Answer 1

Explanation:

y_g = e^(2 x) ( x^2 + 2 x + 1 ) #

Explanation:

Method of Undetermined Coefficients

Start with the homogeneous equation and the complementary solution :

#y'' - 4y' + 4y = 0#

This has characteristic equation:

#lambda^2 - 4lambda + 4 = 0 implies (lambda - 2)^2 = 0#

Repeated roots mean that, in lieu of the usual solution #y_c = alpha e^(lambda_1 x) + beta e^(lambda_2 x)#, we look here for a solution in the form:

#y_c = e^(2 x) ( alpha x + beta )#

And because the non-homogeneous equation already has a #e^(2x)# term, we must look at a particular solution in the form:

#y_p = gamma x^2 e^(2x) #

#implies y' = 2 gamma x e^(2x) + 2 gamma x^2 e^(2x)#

#implies y'' = 2 gamma e^(2x) + 8 gamma x e^(2x) + 4 gamma x^2 e^(2x)#

Putting these into the equation:

# 2 gamma e^(2x) + 8 gamma x e^(2x) + 4 gamma x^2 e^(2x) - 4 (2 gamma x e^(2x) + 2 gamma x^2 e^(2x)) + 4 gamma x^2 e^(2x) = 2 e^(2x)#

#implies gamma = 1 # and #y_p = x^2 e^(2x) #

The general solution is: #y_g = y_c + y_p#

#y_g = e^(2 x) ( alpha x + beta ) + x^2 e^(2x) #

#= e^(2 x) (x^2 + alpha x + beta ) #

Now applying the IV's:

#y(0) = 1 implies beta = 1 implies y_g = e^(2 x) (x^2 + alpha x + 1 ) #

#y' = 2 e^(2 x) (x^2 + alpha x + 1 ) + e^(2 x) (2 x + alpha )#

# = e^(2 x) (2x^2 + (2 alpha + 2) x + (2 + alpha) ) #

And from the second IV, #y'(0) = 4 implies alpha = 2 #

#y_g = e^(2 x) ( x^2 + 2 x + 1 ) #

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