Find the solution to $$ \frac{d y}{d t}=2 y+7 $$ satisfying $$ y(6)=10 $$ $$ y= $$

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We start with the differential equation
$$
\frac{dy}{dt} = 2\,y + 7.
$$
This is a first-order linear differential equation that we can write in standard form as
$$
\frac{dy}{dt} - 2\,y = 7.
$$

**Step 1. Find the integrating factor**

The integrating factor, $$\mu(t)$$, is given by
$$
\mu(t) = e^{\int -2\, dt} = e^{-2t}.
$$

**Step 2. Multiply the differential equation by the integrating factor**

Multiplying both sides by $$e^{-2t}$$ yields
$$
e^{-2t}\frac{dy}{dt} - 2e^{-2t}y = 7e^{-2t}.
$$
The left-hand side is the derivative of $$e^{-2t}y$$, so we have:
$$
\frac{d}{dt}\left(e^{-2t}y\right) = 7e^{-2t}.
$$

**Step 3. Integrate both sides**

Integrate with respect to $$t$$:
$$
\int \frac{d}{dt}\left(e^{-2t}y\right) \, dt = \int 7e^{-2t}\, dt.
$$
This gives
$$
e^{-2t}y = 7\int e^{-2t}\, dt.
$$
The integral on the right is
$$
\int e^{-2t}\, dt = -\frac{1}{2}e^{-2t} + C,
$$
so
$$
e^{-2t}y = 7\left(-\frac{1}{2}e^{-2t}\right) + C = -\frac{7}{2}e^{-2t} + C.
$$

**Step 4. Solve for $$y$$**

Multiply both sides by $$e^{2t}$$:
$$
y = -\frac{7}{2} + Ce^{2t}.
$$

**Step 5. Apply the initial condition**

The initial condition given is $$y(6) = 10$$. Substitute $$t = 6$$ into the general solution:
$$
10 = -\frac{7}{2} + C e^{12}.
$$
Solve for $$C$$:
$$
C e^{12} = 10 + \frac{7}{2} = \frac{20}{2} + \frac{7}{2} = \frac{27}{2}.
$$
Thus,
$$
C = \frac{27}{2} e^{-12}.
$$

Substitute $$C$$ back into the solution:
$$
y = -\frac{7}{2} + \frac{27}{2} e^{-12} e^{2t}.
$$
This simplifies to
$$
y = -\frac{7}{2} + \frac{27}{2} e^{2(t-6)}.
$$

$$
y = -\frac{7}{2} + \frac{27}{2} e^{2(t-6)}
$$
The solution to the differential equation is: $$ y = -\frac{7}{2} + \frac{27}{2} e^{2(t-6)} $$

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