Y''-2y'+y=e^x，y(0)=y'(0)=0. Solve y(2)?
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The short answer is 2e^2. However to reach this solution you need to work a bit. This is a second order ordinary differential equation, which means that the solution will be a function that comprises of two parts, the particular integral and the complementary function. The complementary function is the easiest to find. It is the solution of the differential equation when the left hand side is 0, i.e. y”-2y’+y=0. This you will find by guessing the solution y=e^αx, where α is the constant to be found. By successive differentiation you find that y’=αe^αx and y”=α^2 e^αx. Substitute this in the equation above and you will find: (α^2 -2α +1)e^αx =0. The exponential cannot be 0 therefore the polynomial in the parentheses must be 0, where when you solve that equation you find the double solution α=1. When you have a double solution, your complementary function will be of the form Ae^αx +Βxe^αx. Therefore in this case the complementary function is Ae^x +Bxe^x. Now, the particular integral will be any function that satisfies the original differential equation. Here you basically have to guess which function may be suitable, but it is most likely going to be one that contains an exponential. In fact it is bx^2 e^x, where b is a constant to be found. You find the constant b using the same method as before, that is successive differentiation to find y’ and y”, which are 2xbe^x +bx^2 e^x and 2be^x +4bxe^x +bx^2 e^x respectively. Plug these answers into the original differential equation and you should get 2be^x=e^x which gives b=1/2. Therefore the particular integral is (x^2 e^x)/2. The full solution is the sum of the complementary function and the particular integral so: y(x)=Ae^x +Bxe^x +(x^2 e^x)/2. Taking the boundary condition y=0 gives A=0 and then by differentiating y(x) once and applying the condition y’=0 gives B=0. The solution is therefore simply y(x)=(x^2 e^x)/2 which means that y(2)= 2e^2.
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