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The slope field above is drawn in the ty-plane. Add these two solutions to the slope field: • Sketch the solution with initial condition y (4) = -4. • Sketch the solution with initial condition y (3) = -4. O Some people make an electronic sketch: copy the image into a graphics program, and draw on top of it. Some people make a paper sketch: print the image onto paper, and draw on top of it. Look closely at how each solution behaves as time increases. A This question has often been difficult for students. Feel free to discuss your ideas! SELECT ALL of the following that are true. The solution with initial condition y (3) = -4 increases for t > 3. The solution with initial condition y (4) -4 increases fort > 4. The solution with initial condition y (4) = -4 is an equilibrium solution. The solution with initial condition y (4) = -4 decreases for t > 4. The solution with initial condition y (3) -4 is an equilibrium solution. The solution with initial condition y (3) = -4 decreases for t > 3. %3D

The slope field above is drawn in the ty-plane. Add these two solutions to the slope field: • Sketch the solution with initial condition y (4) = -4. • Sketch the solution with initial condition y (3) = -4. O Some people make an electronic sketch: copy the image into a graphics program, and draw on top of it. Some people make a paper sketch: print the image onto paper, and draw on top of it. Look closely at how each solution behaves as time increases. A This question has often been difficult for students. Feel free to discuss your ideas! SELECT ALL of the following that are true. The solution with initial condition y (3) = -4 increases for t > 3. The solution with initial condition y (4) -4 increases fort > 4. The solution with initial condition y (4) = -4 is an equilibrium solution. The solution with initial condition y (4) = -4 decreases for t > 4. The solution with initial condition y (3) -4 is an equilibrium solution. The solution with initial condition y (3) = -4 decreases for t > 3. %3D

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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-/ ||| || \ーイ The slope field above is drawn in the ty-plane. Add these two solutions to the slope field: • Sketch the solution with initial condition y (4) = -4. • Sketch the solution with initial condition y (3) = –4. O Some people make an electronic sketch: copy the image into a graphics program, and draw on top of it. Some people make a paper sketch: print the image onto paper, and draw on top of it. Look closely at how each solution behaves as time increases. A This question has often been difficult for students. Feel free to discuss your ideas! SELECT ALL of the following that are true. The solution with initial condition y (3) = -4 increases for t > 3. The solution with initial condition y (4) :-4 increases for t > 4. The solution with initial condition y (4) -4 is an equilibrium solution. The solution with initial condition y (4) = -4 decreases for t > 4. The solution with initial condition y (3) = -4 is an equilibrium solution. The solution with initial condition y (3) = -4 decreases for t > 3.
Transcribed Image Text:-/ ||| || \ーイ The slope field above is drawn in the ty-plane. Add these two solutions to the slope field: • Sketch the solution with initial condition y (4) = -4. • Sketch the solution with initial condition y (3) = –4. O Some people make an electronic sketch: copy the image into a graphics program, and draw on top of it. Some people make a paper sketch: print the image onto paper, and draw on top of it. Look closely at how each solution behaves as time increases. A This question has often been difficult for students. Feel free to discuss your ideas! SELECT ALL of the following that are true. The solution with initial condition y (3) = -4 increases for t > 3. The solution with initial condition y (4) :-4 increases for t > 4. The solution with initial condition y (4) -4 is an equilibrium solution. The solution with initial condition y (4) = -4 decreases for t > 4. The solution with initial condition y (3) = -4 is an equilibrium solution. The solution with initial condition y (3) = -4 decreases for t > 3.
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