Use the method of variation of parameters to solve the initial value problem (sort of!)?

y'' - 4y' + 4y = (12x^(2) - 6x)e^(2x), y(0) = 1, y'(0) = 0.





I have actually solved this differential equation... and found solution to be (without substituting initial values etc..) :


y = c_(1)*e^(2x) + c_(2)*xe^(2x) + x^(4)*e^(2x) - x^(3)*e^(2x)


(where '_' means subscript) which I believe is correct.





But the question gives initial values (y(0) = 1, y'(0) = 0) and I'm not sure what to do with those... I'm rather confused... Somehow the initial values give numerical values of c_1 and c_2...? (if that makes any sense!)


If anyone knows...!!!





Thanks very much :)

Use the method of variation of parameters to solve the initial value problem (sort of!)?
C1 = 1, and C2 = - 2





use the IC on the solution......assuming you have the correct solution, and I read it correctly





y(0) = 1 ---%26gt; 1 = c1 * e^0 + c2 * 0 + 0 - 0


1 = c1* 1 + 0, so c1 = 1


to use the other IC, you must take the derivative of your answer, then use x = 0, and


y ' = 0, y = 1 :





y' = 2*c1*e^2x + c2* ( e^2x + 2xe^2x) + ( 4x^3*e^2x + 2x^4*e^2x ) - ( 3x^2*e^2x + 2x^3*e^2x) .... substituting in:





0 = 2*1*1 + c2 ( 1 + 0) + ( 0 + 0 ) - ( 0 + 0 )


thus , 0 = 2 + c2, and c2 = - 2
Reply:y(0)=c_1*e^0+0....





x=0





e^0=1





y(0)=c_(1)=1





for





y'(0)=0





y'=2e^2x+c_(2)[e^2x+2xe^2x]+....





x=0





y'(0)=2+c_(2)[1+0]=0





c_(2)=-2