I need solution to this second order differential equation: y'' - 4y' + 5y = 2xe^(2x)?

Question

Do not use Lagrange method.



I wrote the equation again here:

y'' - 4*y' + 5*y = 2*x*e^(2x)

Answer 1

Find the complementary function by considering the roots of the quadratic m^2 - 4m +5 = 0
Now find the particular integral and add them.

Answer 2

think of of the left-hand-area (LHS) as an operator that transforms y(x) to the function on the main suitable-hand-area (RHS), for this reason 65sin2x. The operations of derivatives would desire to consequence in a sin function. via fact all of us be conscious of that: d/dx sin(Ax) = Acos(Ax) d/dx cos(Ax) = -Asin(Ax) it could be that the answer is of the form: y(x) = Acos(2x) + Bsin(2x) the place A and B are to be desperate to fulfill the differential equation. notice that: y' = -2Asin2x + 2Bcos2x y'' = -4*(Acos2x + Bsin2x) = -4y for this reason: y'' + 4y' + 5y = 4y' + y = 65sin2x to fulfill this equation we would desire to have that: -8Asin2x + 8Bcos2x + Acos2x + Bsin2x = 65sin2x This ends up in 2 equations for our unknowns A and B: 8B + A = 0 -8A + B = sixty 5 with ideas A = -8 and B = . for this reason, the answer is: y(x) = -8cos2x + sin2x

Answer 3

That's more of a statement then a question, as such, I am unable to answear a statement.