Using Second Order Differentiation how do you solve y4y5ex4x

Using Second Order Differentiation, how do you solve y\'\'+4y=5e^x-4x?

Characteristic equation is:

r^2+4 = 0

the roots are:

r = -2i , 2i

So the homogeneous part of the solution is:

yh = Asin(2x) + Bcos(2x)

To find particular solution we have:

p(x) = 5e^x - 4x -> yp = Ce^x + Dx + E

Thus:

yp\' = Ce^x + D

yp\" = Ce^x

and

yp\" + 4yp = 5Ce^x + 4Dx + 4E = 5e^x - 4x

5C = 5 -> C = 1

4D = -4 -> D = -1

4E = 0 -> E = 0

So the final solution is:

y = yh+ yp = Asin(2x) + Bcos(2x) + e^x - x

$Using Second Order Differentiation, how do you solve y\'\'+4y=5e^x-4x?SolutionCharacteristic equation is: r^2+4 = 0 the roots are: r = -2i , 2i So the homogeneo$

Using Second Order Differentiation how do you solve y4y5ex4x

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