non-homogeneous differential equations?!?

Question

how do u get the general solution of y" + 4y' + 3y =x +1

iv got alot of similiar question like this, but i dont know where to begin, if you can help me i would REALLY appreciate it!!!!!

Answer 1

first solve y" + 4y' + 3y = 0; for this there are well known methods that always works.

next find one particular solution :
y" + 4y' + 3y =x +1

you can try a solution that looks like x + 1, for instance y = ax^2 + bx + c, differentiate to get y' and y'', plug into the equations, that should give you values for a , b and c.

Answer 2

There are two solutions that make up the general solution of the general solution. You have to find the homogeneous solution, and then you have to find the nonhomogeneous solution(the particular solution).
For the homogeneous solution, you have to find the characteristic equation along with the characteristic roots. The characteristic equation is: (x^2)+(4x)+3=(x+1)(x+3), so x=-3,x=-1, so the homogeneous solution is y=c*e^(-3x)+d*e^(-x), where c and d are constants to be found from given initial values.

For the particular solution, i don't really know what methods you have learned yet. You could either use undetermined coefficients, or the method of variation of parameters. The method of undetermined coefficients, it will get messy, but all you have to do is assume a solution and put it back in the differential equation to find the coefficients. You would assume a solution of:
y=(A*x)+B+C, but since B+C, is just another constant: y=(A*x)+(D), where A and D are constants to be found once you put the assume solution in the differential equation.

The solution of the differential equation should be y(homogeneous)+y(particular)=Y
Y=c*e^(-3x)+d*e^(-x)+A*x+D, so just make sure your answer is of the form above. I would do the problem but the method of undetermined coefficients is too messy. Good luck! Email me if you have any question.