Explain how to find the general solution to the given ODE, and the particular solution to the given IVP.

\[ 12 \, {y} - 4 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -32 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -4 \, e^{\left(3 \, t\right)} \]

Explain how to find the general solution to the given ODE, and the particular solution to the given IVP.

\[ 9 \, {y} - 3 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -24 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -3 \, e^{\left(3 \, t\right)} \]

Explain how to find the general solution to the given ODE, and the particular solution to the given IVP.

\[ -9 \, {y} = 3 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= -\frac{1}{9} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = -3 \, e^{\left(-3 \, t\right)} \]

\[ 12 \, {y} = -4 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= \frac{1}{4} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = 2 \, e^{\left(-3 \, t\right)} \]

\[ -4 \, {y} = -2 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= 27 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = 3 \, e^{\left(2 \, t\right)} \]

\[ 0 = -4 \, {y'} - 8 \, {y} ,\hspace{1em} y\big( \log\left(3\right) \big)= -\frac{2}{9} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -2 \, e^{\left(-2 \, t\right)} \]

\[ -10 \, {y} = 5 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= \frac{1}{3} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 3 \, e^{\left(-2 \, t\right)} \]

\[ 4 \, {y'} = -12 \, {y} ,\hspace{1em} y\big( \log\left(3\right) \big)= \frac{2}{27} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = 2 \, e^{\left(-3 \, t\right)} \]

\[ 0 = 8 \, {y} + 4 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= -\frac{1}{2} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -2 \, e^{\left(-2 \, t\right)} \]

\[ 6 \, {y} = -3 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= -\frac{2}{9} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -2 \, e^{\left(-2 \, t\right)} \]

\[ -8 \, {y} + 4 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -8 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = -2 \, e^{\left(2 \, t\right)} \]

\[ -3 \, {y'} = -9 \, {y} ,\hspace{1em} y\big( \log\left(3\right) \big)= -108 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -4 \, e^{\left(3 \, t\right)} \]

\[ -4 \, {y} = -2 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= -36 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = -4 \, e^{\left(2 \, t\right)} \]

\[ -8 \, {y} + 4 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= 16 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = 4 \, e^{\left(2 \, t\right)} \]

\[ -4 \, {y'} = 12 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= -\frac{1}{4} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = -2 \, e^{\left(-3 \, t\right)} \]

\[ 12 \, {y} = 4 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= -81 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -3 \, e^{\left(3 \, t\right)} \]

\[ -6 \, {y} = -3 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= 8 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = 2 \, e^{\left(2 \, t\right)} \]

\[ 0 = 2 \, {y'} + 4 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= -1 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -4 \, e^{\left(-2 \, t\right)} \]

\[ 0 = -6 \, {y} - 2 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= \frac{1}{4} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = 2 \, e^{\left(-3 \, t\right)} \]

\[ 3 \, {y'} = 9 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= -16 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -2 \, e^{\left(3 \, t\right)} \]

\[ 4 \, {y'} - 8 \, {y} = 0 ,\hspace{1em} y\big( \log\left(3\right) \big)= -27 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = -3 \, e^{\left(2 \, t\right)} \]

\[ 4 \, {y'} - 12 \, {y} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -16 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -2 \, e^{\left(3 \, t\right)} \]

\[ 0 = 4 \, {y} - 2 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= 12 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = 3 \, e^{\left(2 \, t\right)} \]

\[ 2 \, {y'} = -6 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= \frac{3}{8} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = 3 \, e^{\left(-3 \, t\right)} \]

\[ -3 \, {y'} + 9 \, {y} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -24 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -3 \, e^{\left(3 \, t\right)} \]

\[ 6 \, {y} + 2 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(3\right) \big)= -\frac{4}{27} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = -4 \, e^{\left(-3 \, t\right)} \]

\[ -8 \, {y} = 4 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= \frac{3}{4} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 3 \, e^{\left(-2 \, t\right)} \]

\[ 0 = 5 \, {y'} + 15 \, {y} ,\hspace{1em} y\big( \log\left(3\right) \big)= -\frac{4}{27} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = -4 \, e^{\left(-3 \, t\right)} \]

\[ 5 \, {y'} - 10 \, {y} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= 8 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = 2 \, e^{\left(2 \, t\right)} \]

\[ 2 \, {y'} + 6 \, {y} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -\frac{1}{2} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = -4 \, e^{\left(-3 \, t\right)} \]

\[ 0 = -3 \, {y'} + 6 \, {y} ,\hspace{1em} y\big( \log\left(3\right) \big)= -27 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = -3 \, e^{\left(2 \, t\right)} \]

\[ 5 \, {y'} + 10 \, {y} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -\frac{1}{2} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -2 \, e^{\left(-2 \, t\right)} \]

\[ 0 = -10 \, {y} - 5 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= 1 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 4 \, e^{\left(-2 \, t\right)} \]

\[ 6 \, {y} = -2 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= -\frac{1}{4} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = -2 \, e^{\left(-3 \, t\right)} \]

\[ 0 = 6 \, {y} + 3 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= 1 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 4 \, e^{\left(-2 \, t\right)} \]

\[ -6 \, {y} - 3 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(3\right) \big)= \frac{4}{9} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 4 \, e^{\left(-2 \, t\right)} \]

\[ 0 = 4 \, {y'} - 12 \, {y} ,\hspace{1em} y\big( \log\left(3\right) \big)= -54 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = -2 \, e^{\left(3 \, t\right)} \]

\[ -8 \, {y} + 4 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -12 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = -3 \, e^{\left(2 \, t\right)} \]

\[ 10 \, {y} + 5 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(3\right) \big)= \frac{4}{9} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 4 \, e^{\left(-2 \, t\right)} \]

\[ 0 = 3 \, {y'} + 6 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= \frac{1}{2} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 2 \, e^{\left(-2 \, t\right)} \]

\[ 8 \, {y} = -4 \, {y'} ,\hspace{1em} y\big( \log\left(2\right) \big)= -\frac{1}{2} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -2 \, e^{\left(-2 \, t\right)} \]

\[ 4 \, {y'} = 12 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= 24 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = 3 \, e^{\left(3 \, t\right)} \]

\[ -10 \, {y} = 5 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= \frac{1}{3} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 3 \, e^{\left(-2 \, t\right)} \]

\[ 0 = -5 \, {y'} + 10 \, {y} ,\hspace{1em} y\big( \log\left(3\right) \big)= -36 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(2 \, t\right)} \]

\[ {y} = -4 \, e^{\left(2 \, t\right)} \]

\[ 9 \, {y} = 3 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= 81 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = 3 \, e^{\left(3 \, t\right)} \]

\[ -5 \, {y'} - 10 \, {y} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= -1 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -4 \, e^{\left(-2 \, t\right)} \]

\[ 12 \, {y} - 4 \, {y'} = 0 ,\hspace{1em} y\big( \log\left(2\right) \big)= 32 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(3 \, t\right)} \]

\[ {y} = 4 \, e^{\left(3 \, t\right)} \]

\[ 5 \, {y'} = -15 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= -\frac{3}{8} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-3 \, t\right)} \]

\[ {y} = -3 \, e^{\left(-3 \, t\right)} \]

\[ -10 \, {y} = 5 \, {y'} ,\hspace{1em} y\big( \log\left(3\right) \big)= \frac{2}{9} \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = 2 \, e^{\left(-2 \, t\right)} \]

\[ -4 \, {y'} = 8 \, {y} ,\hspace{1em} y\big( \log\left(2\right) \big)= -1 \]

Then show how to verify that your particular solution is correct.

**Answer:**

\[ {y} = k e^{\left(-2 \, t\right)} \]

\[ {y} = -4 \, e^{\left(-2 \, t\right)} \]