I'm given an equation to solve and I need to use the "method of undetermined coefficients" all I need to find is A,B,C,D I'm not given any initial values so I cant solve the entire thing. This is what I have but I'm not sure if it's correct or not. Thanks for any help... Equation: 2y''+5y'-3y=-2t^3+4t^2 My work: 2r^2+5r-3=0 r=1/2, r=-3 Yh=C1e^(1/2t) + C2e^(-3t) Assume Yp=(At^3+Bt^2+Ct+D) yp'=3At^2-2Bt+C yp''=6At-2B plugging the derivatives into the original equation I get: 12At-4B+15At^2-10Bt+5C-3At^3-3Bt^2-3Ct-3D=-2t^3+4t^2 Solving for A I get: -3At^3=-2t^3 -> A=2/3 Solving for B I get: 10t^2-3Bt^2=4t^2 -> B=2 ? Solving for C I get: 8t-20t-3Ct=0 -> C=-4 ? Solving for D I get:-8-20-3D=0 -> D=-28/3 ? The "?" are the ones I dont know are right. I'm pretty sure the value for A is right though. Also, I plugged in the values for A, B, C, D into the equation as I solved for them, incase you were wondering where some of the numbers came from... I really need help with this, so anything you can do will be greatly appreciated.