<item ident="C2-5395" title="C2 | Non-homogeneous first-order linear ODE | ver. 5395">
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<p>
<strong>C2.</strong>
</p>
<p>Explain how to find the general solution to the given ODE.</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?48 \, e^{\left(-4 \, t\right)} = -12 \, {y} + 3 \, {y'}" alt="48 \, e^{\left(-4 \, t\right)} = -12 \, {y} + 3 \, {y'}" title="48 \, e^{\left(-4 \, t\right)} = -12 \, {y} + 3 \, {y'}" data-latex="48 \, e^{\left(-4 \, t\right)} = -12 \, {y} + 3 \, {y'}"/>
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<p>
<strong>C2.</strong>
</p>
<p>Explain how to find the general solution to the given ODE.</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?48%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D%20=%20-12%20%5C,%20%7By%7D%20+%203%20%5C,%20%7By'%7D" alt="48 \, e^{\left(-4 \, t\right)} = -12 \, {y} + 3 \, {y'}" title="48 \, e^{\left(-4 \, t\right)} = -12 \, {y} + 3 \, {y'}" data-latex="48 \, e^{\left(-4 \, t\right)} = -12 \, {y} + 3 \, {y'}">
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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} = k e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" alt="{y} = k e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" title="{y} = k e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" data-latex="{y} = k e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}"/>
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<h4>Partial Answer:</h4>
<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=%20k%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = k e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" title="{y} = k e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" data-latex="{y} = k e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}">
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