Find the solution of the initial value problem: y (4) -6y +9y"=0 [y(0) = -2, y' (0) = -3, y" (0) = 0, y" (0)=0] a) b) c) d) e) y=-3-2x y=-2-3x 3 y=-3 e³x-2x e³x y = −2+3x 3 y=-2 e³x -3xe³x f) None of the above.

12) Use y = [₁,2"+ to solve 2x²y" - xy + (x+1)y=0. n=0

-2x 8.) y" +2y'+5y= 3e* ; and y(0) = y'(0) = 1

8. Solve (a) y' = (3y²)-1 (Hint: Separate.)

ii) Solve (x D²- 3xD+4) y = x² given that y(1)=1 and y'(1)= 0.

Let y₁ = xe-* be a solution of the equation y" + 2y' + y = 0. Find the second solution from y₁ by reduction of order (apply formula), and then find the general solution of the equation.

5) Form a DE by eliminating the arbitrary constants from the equation y = asin(4x + 1).

Use the method of undetermined coefficients to determine the form of a particular solution for the given equation. y'"' + 10y" – 11y=xe* + 4x What is the form of the particular solution with undetermined coefficients? Yp(x) = ] (Do not use d, D, e, E, i, or I as arbitrary constants since these letters already have defined meanings.)

In Calculus you will see the symbol y'. Treat y' as a variable and solve the equation for y'. 6y²y + 30xy + 6x²y = 5y² + 25xyy' Select one: O a. y = O b. y = = O c. y' = O d.y= = y(y - 6x) 6x² − 5xy + 6y² y - 6x 6x² - 5x + 6y 5y(y – x) x² - 25xy + y² 5y(y - 6x) 6x² - 25xy + 6y²