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ubuntu2004
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<exercise checkit-seed="0001" checkit-slug="AA1" checkit-title="Structure of an IVP and Verifying Solutions">
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<statement>
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<p>
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For each of the following Initial Value Problems (IVPs), designate the following:
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</p>
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<ul>
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<li>its Ordinary Differential Equation (ODE)</li>
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<li>its Initial Value or Values (IVs)</li>
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<li>the order of the IVP</li>
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<li>its independent variable</li>
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<li>its dependent variable</li>
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<li>whether its solution is implicit or explicit</li>
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</ul>
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<p>
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Then show how to verify that its solution is valid.
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</p>
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<ol>
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<li>
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<ul>
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<li>IVP:
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<m>
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8 \, y + 2 \, {\frac{dy}{dt}} - {\frac{d^2y}{dt^2}} = 0;\qquad
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y(0)=
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1,
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y'(0)=
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4
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</m>
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</li>
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<li>Solution: <m>y = e^{\left(4 \, t\right)}</m></li>
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</ul>
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</li>
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<li>
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<ul>
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<li>IVP:
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<m>
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-4 \, t {y'} + 2 \, y {y'} = 4 \, y;\qquad
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y(1)=
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-1 </m>
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</li>
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<li>Solution: <m>-4 \, t y + y^{2} = 5</m></li>
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</ul>
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</li>
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<li>
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<ul>
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<li>IVP:
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<m>
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t {\frac{dx}{dt}} - 8 \, t = 3 \, x;\qquad
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x(1)=
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-1 </m>
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</li>
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<li>Solution: <m>x = 3 \, t^{3} - 4 \, t</m></li>
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</ul>
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</li>
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</ol>
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</statement>
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<answer>
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<ol>
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<li>
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<ul>
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<li>ODE: <m>8 \, y + 2 \, {\frac{dy}{dt}} - {\frac{d^2y}{dt^2}} = 0</m></li>
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<li>IV(s): <m>
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y(0)=1
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,y'(0)=4
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</m></li>
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<li>Order: 2nd</li>
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<li>Independent variable: <m>t</m></li>
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<li>Dependent variable: <m>y</m></li>
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<li>The solution <m>y = e^{\left(4 \, t\right)}</m> is explicit.</li>
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</ul>
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</li>
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<li>
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<ul>
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<li>ODE: <m>-4 \, t {y'} + 2 \, y {y'} = 4 \, y</m></li>
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<li>IV(s): <m>
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y(1)=-1
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</m></li>
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<li>Order: 1st</li>
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<li>Independent variable: <m>t</m></li>
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<li>Dependent variable: <m>y</m></li>
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<li>The solution <m>-4 \, t y + y^{2} = 5</m> is implicit.</li>
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</ul>
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</li>
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<li>
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<ul>
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<li>ODE: <m>t {\frac{dx}{dt}} - 8 \, t = 3 \, x</m></li>
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<li>IV(s): <m>
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x(1)=-1
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</m></li>
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<li>Order: 1st</li>
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<li>Independent variable: <m>t</m></li>
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<li>Dependent variable: <m>x</m></li>
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<li>The solution <m>x = 3 \, t^{3} - 4 \, t</m> is explicit.</li>
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</ul>
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</li>
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</ol>
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</answer>
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</exercise>
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