<item ident="F1-4371" title="F1 | Direction fields for first-order ODEs | ver. 4371">
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<p>
<strong>F1.</strong>
</p>
<p> Use <a href="https://sagecell.sagemath.org/">https://sagecell.sagemath.org/</a> to run the SageMath code <code>t,y = var('t y'); plot_slope_field(sin(y/2), (t,-5,5), (y,-5,5))</code> producing the direction field for the ODE <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y'} = \sin\left(\frac{1}{2} \, {y}\right)" alt="{y'} = \sin\left(\frac{1}{2} \, {y}\right)" title="{y'} = \sin\left(\frac{1}{2} \, {y}\right)" data-latex="{y'} = \sin\left(\frac{1}{2} \, {y}\right)"/>. </p>
<p> Let <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"/> be the solution to the following IVP. Explain how to use its direction field to approximate the value of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"/> at <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?t= -3" alt="t= -3" title="t= -3" data-latex="t= -3"/>. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y'} = \sin\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= -2" alt="{y'} = \sin\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= -2" title="{y'} = \sin\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= -2" data-latex="{y'} = \sin\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= -2"/>
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<p>
<strong>F1.</strong>
</p>
<p> Use <a href="https://sagecell.sagemath.org/">https://sagecell.sagemath.org/</a> to run the SageMath code <code>t,y = var('t y'); plot_slope_field(sin(y/2), (t,-5,5), (y,-5,5))</code> producing the direction field for the ODE <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By'%7D%20=%20%5Csin%5Cleft(%5Cfrac%7B1%7D%7B2%7D%20%5C,%20%7By%7D%5Cright)" alt="{y'} = \sin\left(\frac{1}{2} \, {y}\right)" title="{y'} = \sin\left(\frac{1}{2} \, {y}\right)" data-latex="{y'} = \sin\left(\frac{1}{2} \, {y}\right)">. </p>
<p> Let <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"> be the solution to the following IVP. Explain how to use its direction field to approximate the value of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"> at <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?t=%20-3" alt="t= -3" title="t= -3" data-latex="t= -3">. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By'%7D%20=%20%5Csin%5Cleft(%5Cfrac%7B1%7D%7B2%7D%20%5C,%20%7By%7D%5Cright)%20%5Chspace%7B2em%7D%20y(%20-1%20)=%20-2" alt="{y'} = \sin\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= -2" title="{y'} = \sin\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= -2" data-latex="{y'} = \sin\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= -2">
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<h4>Partial Answer:</h4>
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<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p( -3 )\approx -2.0" alt="y_p( -3 )\approx -2.0" title="y_p( -3 )\approx -2.0" data-latex="y_p( -3 )\approx -2.0"/>
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<h4>Partial Answer:</h4>
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<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p(%20-3%20)%5Capprox%20-2.0" alt="y_p( -3 )\approx -2.0" title="y_p( -3 )\approx -2.0" data-latex="y_p( -3 )\approx -2.0">
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