(a) The general solution of the differential equation y\' = y + 2t is a family o
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Question
(a) The general solution of the differential equation y' = y + 2t is a family of curves. Given any point in the (t, y) plane one member of this family of curves passes through this point. Consider the solution curve to this differential equation that passes through the particular point (1, 2). What is the slope of the tangent line to this curve at the point (1, 2)? (b) Consider the particular solution of y' = y +2t which passes through the point (2, -1). Suppose we find the tangent line to this curve at the point (2, -1) which we know is on this curve. What is the slope of this tangent line? (c) Suppose we find the tangent line to the solution curve at each of the following points. What is the slope of this tangent line? (0,2) (2,1) (1,0) (-1,2)(-2,1)(-2,-2) (d) Draw a short segment of each of these tangent lines on a ty axis.Explanation / Answer
(a) slope of tangent at (1,2) = y'(1,2) = 2+2(1) = 4... (b) (2,-1) ... slope = -1+2(2) = 3... equation is (y+1)=3(t-2) ==> 3t-y-7 = 0.... (c) (0,2) slope = 2+ 2(0) =2..... (2,1) slope= 1+2(2) = 5.......... (1,0) slope = 0+2(1) = 2........... (-1,2) slope = 2+2(-1) = 0........... (-2,1) slope = 1+2(-2) = -3....... (-2,-2) slope = -2+2(-2) =-6............. (d) like we calculated in (b) part the equations can be calculated for all these points and the graph would be trivial from there......
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