As we have seen, a solution of t2y''+6ty′+4y=28t3+8 can be found by writing yp=u1y1+u2y2 ​as done in lecture. Find u1u1​.

Question

As we have seen, a solution of t²y''+6ty′+4y=28t³+8 can be found by writing yp=u1y1+u2y2 ​as done in lecture. Find u1u1​.

−4t⁷/3−2t⁴/3

4t⁷/3+2t⁴/3

4t⁷/3+2t⁴/3

−7t⁴/3−8t/3

Answer 1

I can show you how to find the general solution to the equation where the right hand side equals 0 and a specific solution for the given right hand side.

Given the form of this differential equation, we can consider solutions of the form c1t^n1 + c2t^n2

We solve for n as follows: as the first derivative of t^n = nt^n-1, and the second derivative of t^n = n(n-1)t^n-2, we get for this equation

n(n-1) + 6(n) + 4 = 0

n^2 - n + 6n + 4 = 0

n^2 + 5n + 4 = 0

(n + 4)(n + 1) = 0

n = -4, -1

The general solution is c1t^-1 + c2t^-4 or c1/t + c2/t^4

Next, consider the specific solution.

For the 28t^3 term, we get

c(3(3-1) + 6(3) + 4) = 28

c(6 + 18 + 4) = 28

28c = 28

c = 1

From the 8 term, we get

4c = 8

c = 2

Our specific solution is t^3 + 2

Thus, our general solution is t^3 + 2 + c1/t + c2/t^4

I hope this helps with the question that you are asked.