Generating 1 private exercises for CC1... Done!
Explain how to find the general solution to the given ODE, and the particular solution that satisfies the given initial value.
−6y=3y′,y(log(2))=−43
Answer:
y=ke(−2t)
y=−3e(−2t)
Data XML
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<data seed="3335">
<ode>-6 \, {y} = 3 \, {y'}</ode>
<ode_sol>{y} = k e^{\left(-2 \, t\right)}</ode_sol>
<ivp_sol>{y} = -3 \, e^{\left(-2 \, t\right)}</ivp_sol>
<y0>-\frac{3}{4}</y0>
<t0>\log\left(2\right)</t0>
</data>
HTML source
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<div class="checkit exercise" data-checkit-slug="CC1" data-checkit-title="Constant-Coefficient Linear Homogeneous First-Order IVPs" data-checkit-seed="3335">
<div class="exercise-statement">
<p> Explain how to find the general solution to the given ODE, and the particular solution that satisfies the given initial value. </p>
<p class="math math-display">\[-6 \, {y} = 3 \, {y'},\hspace{1em}
y\big(\log\left(2\right)\big)=-\frac{3}{4}\]</p>
</div>
<div class="exercise-answer">
<p>
<b>Answer:</b>
</p>
<p class="math math-display">\[{y} = k e^{\left(-2 \, t\right)}\]</p>
<p class="math math-display">\[{y} = -3 \, e^{\left(-2 \, t\right)}\]</p>
</div>
</div>
LaTeX source
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\begin{exercise}{CC1}{Constant-Coefficient Linear Homogeneous First-Order IVPs}{3335}
\begin{exerciseStatement}
Explain how to find the general solution to the given ODE, and the particular solution that satisfies the given initial value.
\[-6 \, {y} = 3 \, {y'},\hspace{1em}
y\big(\log\left(2\right)\big)=-\frac{3}{4}\]\end{exerciseStatement}
\begin{exerciseAnswer}\[{y} = k e^{\left(-2 \, t\right)}\]\[{y} = -3 \, e^{\left(-2 \, t\right)}\]\end{exerciseAnswer}
\end{exercise}
QTI source
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<p> Explain how to find the general solution to the given ODE, and the particular solution that satisfies the given initial value. </p>
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<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://usaonline.southalabama.edu/equation_images/-6 \, {y} = 3 \, {y'},\hspace{1em} y\big(\log\left(2\right)\big)=-\frac{3}{4}" alt="-6 \, {y} = 3 \, {y'},\hspace{1em} y\big(\log\left(2\right)\big)=-\frac{3}{4}" title="-6 \, {y} = 3 \, {y'},\hspace{1em} y\big(\log\left(2\right)\big)=-\frac{3}{4}" data-latex="-6 \, {y} = 3 \, {y'},\hspace{1em} y\big(\log\left(2\right)\big)=-\frac{3}{4}"/>
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<strong>CC1.</strong>
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<p> Explain how to find the general solution to the given ODE, and the particular solution that satisfies the given initial value. </p>
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<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://usaonline.southalabama.edu/equation_images/-6%20%5C,%20%7By%7D%20=%203%20%5C,%20%7By'%7D,%5Chspace%7B1em%7D%20y%5Cbig(%5Clog%5Cleft(2%5Cright)%5Cbig)=-%5Cfrac%7B3%7D%7B4%7D" alt="-6 \, {y} = 3 \, {y'},\hspace{1em} y\big(\log\left(2\right)\big)=-\frac{3}{4}" title="-6 \, {y} = 3 \, {y'},\hspace{1em} y\big(\log\left(2\right)\big)=-\frac{3}{4}" data-latex="-6 \, {y} = 3 \, {y'},\hspace{1em} y\big(\log\left(2\right)\big)=-\frac{3}{4}">
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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://usaonline.southalabama.edu/equation_images/%7By%7D%20=%20k%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = k e^{\left(-2 \, t\right)}" title="{y} = k e^{\left(-2 \, t\right)}" data-latex="{y} = k e^{\left(-2 \, t\right)}">
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PreTeXt source
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<exercise checkit-seed="3335" checkit-slug="CC1" checkit-title="Constant-Coefficient Linear Homogeneous First-Order IVPs">
<statement>
<p>
Explain how to find the general solution to the given ODE,
and the particular solution that satisfies the given initial value.
</p>
<me>-6 \, {y} = 3 \, {y'},\hspace{1em}
y\big(\log\left(2\right)\big)=-\frac{3}{4}</me>
</statement>
<answer>
<me>{y} = k e^{\left(-2 \, t\right)}</me>
<me>{y} = -3 \, e^{\left(-2 \, t\right)}</me>
</answer>
</exercise>