Hello, I have a midterm this afternoon in D.E. I tried to do this problem over the weekend but I cannot get the final answer:
I have this second order equation
x<sup>2</sup>y" - 2xy' + 2y = x<sup>2</sup> + 2
for which I am trying to solve the Particular solution for. However, I cannot set y<sub>p</sub> = Ax<sup>2</sup>+Bx+C because this is a second-order Euler-Cauchy differential equation. Instead, I must set y<sub>p</sub> = Au*e<sup>2u</sup>+B where u = ln(x), and x = e<sup>u</sup>.
After I plug the Particular solution back into the D.E., I don't get how the answer is A =1 and B =1. I think the problem I am having is grouping the e<sup>u</sup> terms and equating them to the right hand side.
This is what I get after plugging y<sub>p</sub> = e<sup>4u</sup>[4Au+4A] + e<sup>3u</sup>[-4Au-2A]+e<sup>2u</sup>[2Au]+2B = e<sup>2u</sup>+2
I set 2Au = 1, and 2B = 2. Can someone please verify this? Thanks.
I have this second order equation
x<sup>2</sup>y" - 2xy' + 2y = x<sup>2</sup> + 2
for which I am trying to solve the Particular solution for. However, I cannot set y<sub>p</sub> = Ax<sup>2</sup>+Bx+C because this is a second-order Euler-Cauchy differential equation. Instead, I must set y<sub>p</sub> = Au*e<sup>2u</sup>+B where u = ln(x), and x = e<sup>u</sup>.
After I plug the Particular solution back into the D.E., I don't get how the answer is A =1 and B =1. I think the problem I am having is grouping the e<sup>u</sup> terms and equating them to the right hand side.
This is what I get after plugging y<sub>p</sub> = e<sup>4u</sup>[4Au+4A] + e<sup>3u</sup>[-4Au-2A]+e<sup>2u</sup>[2Au]+2B = e<sup>2u</sup>+2
I set 2Au = 1, and 2B = 2. Can someone please verify this? Thanks.