Another problem... how do I solve this one?
y'' − 3y' + 2y = − 6e^{4x}
y(0) = 1
y'(0) = 1
x = 1
I think I have to do the left part first (y'' − 3y' + 2y = 0) and then the right part of the equation (I don't now how to solve this one).
Where do I put in y(0) and y'(0)?
y'' − 3y' + 2y = − 6e^{4x}
y(0) = 1
y'(0) = 1
x = 1
I think I have to do the left part first (y'' − 3y' + 2y = 0) and then the right part of the equation (I don't now how to solve this one).
Where do I put in y(0) and y'(0)?