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1. The direction field for dy/dx = 4x/y is shown in Figure 1.12. (a) Verify that the straight lines y = ±2x are solution curves, provided x ? 0. (b) Sketch the solution curve with initial condition y(0) = 2. (c) Sketch the solution curve with initial condition y(2) = 1. (d) What can you say about the behavior of the above solutions as x ? +?? How about x ? -??

          1. The direction field for dy/dx = 4x/y is shown in Figure 1.12. (a) Verify that the straight lines y = ±2x are solution curves, provided x ? 0. (b) Sketch the solution curve with initial condition y(0) = 2. (c) Sketch the solution curve with initial condition y(2) = 1. (d) What can you say about the behavior of the above solutions as x ? +?? How about x ? -??
        
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1. The direction field for dy/dx = 4x/y is shown in Figure 1.12. (a) Verify that the straight lines y = ±2x are solution curves, provided x ? 0. (b) Sketch the solution curve with initial condition y(0) = 2. (c) Sketch the solution curve with initial condition y(2) = 1. (d) What can you say about the behavior of the above solutions as x ? +?? How about x ? -??

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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The direction field for dy/dx = 4x/y is shown in Figure 1.12. (a) Verify that the straight lines y = ±2x are solution curves, provided x ≠ 0. (b) Sketch the solution curve with initial condition y(0) = 2. (c) Sketch the solution curve with initial condition y(2) = 1. (d) What can you say about the behavior of the above solutions as x → +∞? How about x → -∞?
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Consider the differential equation dy/dx = (y - y^2)/x for all x ≠ 0. (a) Verify that y = x/(x + C), x ≠ -C and C ≠ 0 is a general solution for the given differential equation and show that all solutions contain (0, 0). (b) Write an equation of the particular solution that contains the point (1, 2), and find the value of dy/dx at (0, 0) for this solution. (c) Write an equation of the vertical and horizontal asymptotes of the particular solution found in (b). (d) The slope field for the given differential equation is provided. Sketch both branches of the particular solution curve that passes through the point (1, 2).

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Transcript

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00:01 So we have differential equation divided by d x is equals to 4x divided by y.
00:12 If we solve this, we have y, dy is equal to 4x d x integrating we get y squared divided by 2 is equal to 4x squared divided by 2 plus t.
00:29 From this we get y square is equals to 4 x square plus c now if we put when c is equals to 0 we get y square is equal to 4x square or y is equals to plus minus 2x so from this we can write straight line y equals to plus minus 2x is solution...
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