Key Concepts and Formulas

• First-Order Differential Equation: An equation involving a function and its first derivative. The general form we'll aim for is $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (I.F.): For a linear differential equation in the form $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is $I.F. = e^{\int P(x) dx}$. Multiplying the differential equation by the I.F. makes the left-hand side a perfect derivative.
• Solution using Integrating Factor: After multiplying the equation by the I.F., we integrate both sides with respect to $x$ to get the general solution.

Step-by-Step Solution

Step 1: Rewrite the given differential equation.

We are given the differential equation: $2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0$ Our goal is to rearrange this into the form $\frac{dy}{dx} + P(x)y = Q(x)$. First, move the $dy$ term to the right side: $2(3+y) e^{2 x} d x = \left(7+e^{2 x}\right) d y$ Now, divide both sides by $dx$ and $\left(7+e^{2 x}\right)$: $\frac{dy}{dx} = \frac{2(3+y) e^{2 x}}{7+e^{2 x}}$

Step 2: Rearrange to the standard form.

Separate $y$ and non-$y$ terms on the right-hand side: $\frac{dy}{dx} = \frac{6e^{2x} + 2ye^{2x}}{7+e^{2x}}$ $\frac{dy}{dx} - \frac{2e^{2x}}{7+e^{2x}}y = \frac{6e^{2x}}{7+e^{2x}}$ Now the equation is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\frac{2e^{2x}}{7+e^{2x}}$ and $Q(x) = \frac{6e^{2x}}{7+e^{2x}}$.

Step 3: Calculate the Integrating Factor.

The integrating factor is given by $I.F. = e^{\int P(x) dx}$. Therefore, $I.F. = e^{\int -\frac{2e^{2x}}{7+e^{2x}} dx}$ Let $u = 7+e^{2x}$, then $du = 2e^{2x} dx$. So, the integral becomes: $I.F. = e^{-\int \frac{1}{u} du} = e^{-\ln |u|} = e^{\ln |u^{-1}|} = \frac{1}{u} = \frac{1}{7+e^{2x}}$

Step 4: Multiply the differential equation by the Integrating Factor.

Multiply the equation $\frac{dy}{dx} - \frac{2e^{2x}}{7+e^{2x}}y = \frac{6e^{2x}}{7+e^{2x}}$ by $I.F. = \frac{1}{7+e^{2x}}$: $\frac{1}{7+e^{2x}}\frac{dy}{dx} - \frac{2e^{2x}}{(7+e^{2x})^2}y = \frac{6e^{2x}}{(7+e^{2x})^2}$ The left side is now the derivative of the product $y \cdot (I.F.)$. That is, $\frac{d}{dx}\left(y \cdot \frac{1}{7+e^{2x}}\right) = \frac{6e^{2x}}{(7+e^{2x})^2}$

Step 5: Integrate both sides with respect to x.

Integrate both sides with respect to $x$: $\int \frac{d}{dx}\left(y \cdot \frac{1}{7+e^{2x}}\right) dx = \int \frac{6e^{2x}}{(7+e^{2x})^2} dx$ $y \cdot \frac{1}{7+e^{2x}} = \int \frac{6e^{2x}}{(7+e^{2x})^2} dx$ Again, let $u = 7+e^{2x}$, so $du = 2e^{2x}dx$. Then the integral on the right becomes: $\int \frac{3}{u^2} du = 3\int u^{-2} du = 3\left(-\frac{1}{u}\right) + C = -\frac{3}{7+e^{2x}} + C$ So, we have: $\frac{y}{7+e^{2x}} = -\frac{3}{7+e^{2x}} + C$

Step 6: Solve for y.

Multiply both sides by $(7+e^{2x})$: $y = -3 + C(7+e^{2x})$

Step 7: Use the point (0, 5) to find C.

The curve passes through $(0, 5)$, so when $x=0$, $y=5$: $5 = -3 + C(7+e^{2(0)})$ $5 = -3 + C(7+1)$ $8 = 8C$ $C = 1$

Step 8: Write the equation of the curve.

Substitute $C=1$ into the equation: $y = -3 + 1(7+e^{2x}) = 4 + e^{2x}$

Step 9: Use the point $(\log_e 2, k)$ to find k.

The curve passes through $(\log_e 2, k)$, so when $x=\log_e 2$, $y=k$: $k = 4 + e^{2(\log_e 2)} = 4 + e^{\log_e 2^2} = 4 + e^{\log_e 4} = 4 + 4 = 8$ However, the correct answer is 32. Let's reexamine our work. The mistake is in Step 7, where C=1, let's see.

Step 7 (Corrected): Use the point (0, 5) to find C.

The curve passes through $(0, 5)$, so when $x=0$, $y=5$: $5 = -3 + C(7+e^{2(0)})$ $5 = -3 + C(7+1)$ $8 = 8C$ $C = 1$

Step 8 (Corrected): Write the equation of the curve.

Substitute $C=1$ into the equation: $y = -3 + 1(7+e^{2x}) = 4 + e^{2x}$

Step 9 (Corrected): Use the point $(\log_e 2, k)$ to find k.

The curve passes through $(\log_e 2, k)$, so when $x=\log_e 2$, $y=k$: $k = 4 + e^{2(\log_e 2)} = 4 + e^{\log_e 2^2} = 4 + e^{\log_e 4} = 4 + 4 = 8$

The provided answer is 32, but we keep getting 8. Let's re-examine the integration step. $\int \frac{6e^{2x}}{(7+e^{2x})^2} dx$ Let $u = 7 + e^{2x}$, $du = 2e^{2x}dx$ $\int \frac{3}{u^2} du = -\frac{3}{u} + C = -\frac{3}{7+e^{2x}} + C$

So, $\frac{y}{7+e^{2x}} = \frac{-3}{7+e^{2x}} + C$ $y = -3 + C(7+e^{2x})$ Plug in $(0,5)$. $5 = -3 + C(7+1) = -3 + 8C$ $8 = 8C$, $C=1$. $y = -3 + (7+e^{2x}) = 4 + e^{2x}$. Plug in $(\log_e 2, k)$. $k = 4 + e^{2\log_e 2} = 4 + e^{\log_e 4} = 4+4=8$

We are still getting 8. We need to go back and check the original equation. $2(3+y)e^{2x}dx - (7+e^{2x})dy = 0$. $(7+e^{2x})dy = 2(3+y)e^{2x}dx$ $\frac{dy}{dx} = \frac{2(3+y)e^{2x}}{7+e^{2x}}$ $\frac{dy}{dx} = \frac{6e^{2x}+2ye^{2x}}{7+e^{2x}}$ $\frac{dy}{dx} - \frac{2e^{2x}}{7+e^{2x}}y = \frac{6e^{2x}}{7+e^{2x}}$. $P(x) = \frac{-2e^{2x}}{7+e^{2x}}$ $I.F. = e^{\int P(x)dx} = e^{\int \frac{-2e^{2x}}{7+e^{2x}} dx}$ Let $u = 7+e^{2x}, du = 2e^{2x}dx$. $I.F. = e^{\int \frac{-1}{u}du} = e^{-\ln u} = e^{\ln u^{-1}} = \frac{1}{u} = \frac{1}{7+e^{2x}}$. Multiply the diff eq by I.F. $\frac{1}{7+e^{2x}} \frac{dy}{dx} - \frac{2e^{2x}}{(7+e^{2x})^2}y = \frac{6e^{2x}}{(7+e^{2x})^2}$ $\frac{d}{dx}(\frac{y}{7+e^{2x}}) = \frac{6e^{2x}}{(7+e^{2x})^2}$ $\int \frac{d}{dx}(\frac{y}{7+e^{2x}})dx = \int \frac{6e^{2x}}{(7+e^{2x})^2}dx$ $\frac{y}{7+e^{2x}} = \int \frac{6e^{2x}}{(7+e^{2x})^2}dx$ Let $u = 7+e^{2x}, du = 2e^{2x}dx$ $\frac{y}{7+e^{2x}} = \int \frac{3}{u^2} du = \frac{-3}{u} + C = \frac{-3}{7+e^{2x}} + C$. $y = -3 + C(7+e^{2x})$ $(0,5): 5 = -3 + C(7+1)$. $8 = 8C, C=1$. $y = -3 + (7+e^{2x}) = 4 + e^{2x}$ $(\log_e 2, k): k = 4 + e^{2\log_e 2} = 4 + e^{\log_e 4} = 4+4=8$

There must be an error in the question or the answer. Let's assume the correct answer is 8.

Common Mistakes & Tips

• Always double-check your integration, especially when using substitution.
• Be careful with signs when calculating the integrating factor and applying it to the differential equation.
• Verify your solution by plugging it back into the original differential equation.

Summary

We solved the given first-order differential equation by finding the integrating factor, multiplying the equation by the integrating factor, integrating both sides, and then using the given points to find the constants. We arrived at the equation $y = 4 + e^{2x}$ and found $k=8$. However, since the problem states that the correct answer is 32, there may be an issue with the question or given answer.

Final Answer

The final answer is \boxed{8}, which does not correspond to any of the provided options.