SOLVE THE FOLLOWING DE:

Question

y''+7y'+10y=4xe^{-3x} HERE y(0)=0, y'(0)=-1

Answer 1

first ... find the roots of the complementary equation r²+7r+10=0 to be -2, -5. Thus y_c = Ae^-2x+Be^-5x.

next ... note that g(x)=4xe^-3x and -3 is not repeated in the complementary solution. Thus the particular solution will take the form of y_p=(Cx+D)e^-3x. (refer to your text for an assortment of g(x) forms)

next ... find the first and second derivatives of y_p and insert them into the original problem.

next ... equate the left side to the right side to solve for the constants C and D by mapping the results

next ... assemble the final solution by forming y(t) = y_c +y_p

next ... use the intial conditions to solve for A and B to complete the solution.