Particular Solution to Differential Equation

[robvalue]

Been a while here too, haha!

If x(t) = 2 so it's just a constant, that gives

5y' + 10y = 4

So the particular solution would be found by y=c as higher powers of t would end up with zero coefficient, giving

5*0 + 10*c = 4

Then c = 0.4

Solving the homogenous equation

5y' + 10y = 0

y' + 2y = 0

Needs multiplying by the integrating factor of e^(integral of coefficient of y) = e^(2t)

y'e^(2t) + 2ye^(2t) = 0

(ye^(2t))' = 0

ye^(2t) = A

y = Ae^(-2t)

So complete solution is y = Ae^(-2t) + 0.4

If my memory serves me correctly!