Find the general solution of each of the following equations: y\" + y\' - 6y = 0

Question

Find the general solution of each of the following equations: y" + y' - 6y = 0; y" - 6y' + 25y = 0; y" + 2y' + y = 0; 4y" + 20y' + 25y = 0: y' + 8y = 0; y" + 2y' + 3y = 0; 2y" - 4y' + 8y - 0; y" = 4y; y" - 4y' + 4y = 0; y" - 4y' + 4y = 0; y" - 4y' + 4y = 0; y" - 9y' + 20y = 0; 2y" + y' - y = 0; 2y" + 2y' + 3y = 0; 16y" - 8y' + y = 0; 4y" - 12y' + 9y = 0; v" + 4y' + 5y = 0; y" + y' = 0; y" + 4y' - 5y = 0. Find the solutions of the following initial value problems: y'' - 5y' + 6y = 0, y(1) = e^2 and y' (1) = 3e^2; y'' - 6y' + 5y = 0, y(0) = 3 and y' (0) = 11; y'' - 6y' + 9y = 0, y(0) = 0 and y' (0) = 5; y'' + 4y' + 5y = 0, y(0) = 1 and y' (0) = 0; y'' + 4y' + 2y = 0, y(0) = -1 and y'(0) = 2 + 3 squareroot 2; y'' + 8y' - 9y = 0, y(1) = 2 and y' (1) = 0. Show that the general solution of equation (1) approaches 0 as only if p and q are both positive. Without using the formulas obtained in this section, show that the derivative of any solution of equation (1) is also a solution.

Explanation / Answer

Given that

y'' - 4y' + 4y = 0

The auxialary equation is ,

r2 - 4r + 4 = 0

r2 - 2r - 2r + 4 = 0

r(r - 2) -2(r - 2) = 0

(r - 2)(r - 2) = 0

r - 2 = 0 , r - 2 = 0

r = 2 , r = 2

If the roots are real and equal then the solution is ,

y(x) = c1erx + c2xerx

y(x) = c1e2x + c2xe2x

Therefore ,

The general solution is ,

  y(x) = c1e2x + c2xe2x

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