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Jan. 31, 2023•0 likes•4 views

Jan. 31, 2023•0 likes•4 views

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Q1 Q2 Q3 Express the plane that contains the points A = ( - 5, - 2,3), B = (1,1, - 4) and C = ( - 1, - 4,3) in equational form. Your answer should have the form ax + by + cz = d where you give the specific numeric values for a, b, c and d. An equational form is {(x, y, z) R3 | } The Initial Value Problem: y\" -6y\' = 0 with y\' = 16 and y\' = 48 when x = 0, has general solution y = c1e6x + c2. Give the solution to the IVP. The solution to the Initial value problem is y = The Initial Value Problem: y \" + 3 y\' - 10 y = -2 with y = - 4 and y\' = - 1 when x = 0, has associated homogeneous equation y\" + 3 y\' - 10 y = 0 and this has general solution y= c1 e -5x +c2 e 2x. Write down what you would try for a particular solution. Use, as needed, the Letters A, B and C {in that order) as unknown constants and express any polymer powers only. For the particular solution, try y = Solution 2.) r^2 - 6r = 0 so r1 =0 and r2 = 6 y = c1e^6x + c2 e^0x = c1e^6x + c2 to find c1,c2 16 = c1 + c2 and 48 = 6c1 so c1 = 8 c2 = 8 so y = 8e^6x + 8 .

- 1. Q1 Q2 Q3 Express the plane that contains the points A = ( - 5, - 2,3), B = (1,1, - 4) and C = ( - 1, - 4,3) in equational form. Your answer should have the form ax + by + cz = d where you give the specific numeric values for a, b, c and d. An equational form is {(x, y, z) R3 | } The Initial Value Problem: y" -6y' = 0 with y' = 16 and y' = 48 when x = 0, has general solution y = c1e6x + c2. Give the solution to the IVP. The solution to the Initial value problem is y = The Initial Value Problem: y " + 3 y' - 10 y = -2 with y = - 4 and y' = - 1 when x = 0, has associated homogeneous equation y" + 3 y' - 10 y = 0 and this has general solution y= c1 e -5x +c2 e 2x. Write down what you would try for a particular solution. Use, as needed, the Letters A, B and C {in that order) as unknown constants and express any polymer powers only. For the particular solution, try y = Solution
- 2. 2.) r^2 - 6r = 0 so r1 =0 and r2 = 6 y = c1e^6x + c2 e^0x = c1e^6x + c2 to find c1,c2 16 = c1 + c2 and 48 = 6c1 so c1 = 8 c2 = 8 so y = 8e^6x + 8