Solve the IVP: y" - 2y' + y = 0 , y(0)=5 , y'(0)=10. Find y(1)

Question

Solve the IVP: y" - 2y' + y = 0 ,  y(0)=5 , y'(0)=10. Find y(1)

across · Answer

We are told that$$y''-2y'+y=0,$$and$$y(0)=5,$$$$y'(0)=10.$$We make a lambda substitution, we obtain$$λ^2-2λ+1=0.$$Solving this polynomial, we get$$(λ-1)^2=0,$$which is a repeated root at$$λ=1.$$Therefore, the solution to this higher-order ODE is in the form$$y=c_{1}e^{x}+c_{2}xe^{x}.$$I will leave the finding of the constants up to you.

anonymous · Answer

yes, this is exactly what i get everytime. and for C1, I get 5. C2 is 5. and then for for y(1) I get 0. which I am told is the wrong answer. . . HELP!

phi · Answer

your constants look ok, and  y(1) = 5 e^1 + 5*e^1= 10e

anonymous · Answer

Sorry my previous post meant to say that for C2 I am getting -5 after integrating xe^x by parts. which is how I come up with 0 as an answer (which is incorrect). I am getting 10=5e^0+C2(0e^0-e^0). Any more help?

across · Answer

$$y=c_{1}e^{x}+c_{2}xe^{x}.$$$$5=c_{1}e^{0}+c_{2}(0)e^{0}\implies c_{1}=5.$$$$10=c_{1}e^{0}+c_{2}e^{0}+c_{2}(0)e^{0}\implies c_{2}=5.$$$$y=5e^{x}+5xe^{x}.$$

across · Answer

$$y(1)=5e^{1}+5(1)e^{1}=5e+5e=10e,$$as phi already mentioned.