Proof: Solution to dy/dx = sin y + 1, y(0) = -π/6

Answer

To prove the solution, we separate the variables to obtain the integral $\int \frac{1}{\sin y + 1} dy = \int dx$ and utilize the $t = \tan\left(\frac{y}{2}\right)$ substitution. Integrating both sides and applying the initial condition $(0, -\pi/6)$ allows us to solve for the constant of integration and rearrange the expression into the required form.

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Observation of the Image

The image displays a first-order differential equation $\frac{dy}{dx} = \sin y + 1$. Below the equation is a Cartesian plane showing a smooth, increasing curve (an S-shaped sigmoidal curve) that passes through the fourth quadrant at $x=0$, consistent with the initial condition $(0, -\pi/6)$. The curve appears to have horizontal asymptotes, which occur where $\frac{dy}{dx} = 0$.

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Proof

Given:

1. $\frac{dy}{dx} = \sin y + 1$
2. Initial condition: $y(0) = -\frac{\pi}{6}$

To Prove: $y = \frac{\pi}{2} + 2 \tan^{-1}(x - \sqrt{3})$

Proof:

1. Separation of variables We rearrange the differential equation to group terms involving $y$ on one side and terms involving $x$ on the other. $\frac{1}{\sin y + 1} dy = dx$ This setup allows us to integrate each side with respect to its own variable.

2. Application of the Weierstrass Substitution To integrate the left side, we use the $t$-substitution: Let $t = \tan\left(\frac{y}{2}\right)$. From this, we know $\sin y = \frac{2t}{1+t^2}$ and $dy = \frac{2}{1+t^2} dt$. $\int \frac{1}{\frac{2t}{1+t^2} + 1} \cdot \frac{2}{1+t^2} dt = \int dx$ This substitution converts trigonometric integrals into rational function integrals.

3. Simplification of the integral Expanding the denominator on the left: $\int \frac{2}{2t + 1 + t^2} dt = \int dx$ The denominator is a perfect square. $\int \frac{2}{(t+1)^2} dt = \int dx$ We simplified the rational expression to a standard power rule form. ⚠️ This step is required on exams to show clear algebraic manipulation.

4. Integration Perform the integration on both sides: $-2(t+1)^{-1} = x + C$ Substituting $t = \tan\left(\frac{y}{2}\right)$ back in: $\frac{-2}{\tan(\frac{y}{2}) + 1} = x + C$ The constant $C$ represents the family of solutions for the differential equation.

5. Solving for the constant $C$ Apply the point $(0, -\frac{\pi}{6})$. This means when $x = 0$, $y = -\frac{\pi}{6}$. $\frac{-2}{\tan(-\frac{\pi}{12}) + 1} = 0 + C$ Using the exact value $\tan(-\frac{\pi}{12}) = \tan(-15^\circ) = -(2 - \sqrt{3}) = \sqrt{3} - 2$: $C = \frac{-2}{(\sqrt{3} - 2) + 1} = \frac{-2}{\sqrt{3} - 1}$ Rationalizing the denominator: $C = \frac{-2(\sqrt{3}+1)}{3-1} = -(\sqrt{3} + 1)$ The constant defines the specific curve that passes through the given point.

6. Isolating $y$ Substitute $C$ back into the equation: $\frac{-2}{\tan(\frac{y}{2}) + 1} = x - \sqrt{3} - 1$ Multiply both sides by $-1$ and take the reciprocal: $\frac{\tan(\frac{y}{2}) + 1}{2} = \frac{1}{1 + \sqrt{3} - x}$ (Note: For the specific form in the proof, we manipulate the expression for $t$ instead): $\tan\left(\frac{y}{2}\right) + 1 = \frac{-2}{x - \sqrt{3} - 1}$ To match the required inverse tangent form, we rearrange the integrated form $-2(t+1)^{-1} = x - \sqrt{3} - 1$: $(t+1) = \frac{-2}{x - \sqrt{3} - 1} = \frac{2}{1 + \sqrt{3} - x}$ However, the target expression $y = \frac{\pi}{2} + 2 \tan^{-1}(x - \sqrt{3})$ implies a different algebraic path. Let's verify the target solution satisfies the D.E.: If $y = \frac{\pi}{2} + 2 \tan^{-1}(x - \sqrt{3})$, then $\tan(\frac{y}{2} - \frac{\pi}{4}) = x - \sqrt{3}$. Using the identity $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$: $\frac{\tan(\frac{y}{2}) - 1}{\tan(\frac{y}{2}) + 1} = x - \sqrt{3}$ This confirms the relationship between $x$ and $y$ derived from the integral.

7. Final Rearrangement From the integral step: $\frac{-2}{t+1} = x + C$. With $C = -\sqrt{3}-1$: $\frac{2}{t+1} = \sqrt{3} + 1 - x$ Using the identity $\frac{1 - t}{1 + t} = \tan(\frac{\pi}{4} - \frac{y}{2})$: $\tan\left(\frac{y}{2} - \frac{\pi}{4}\right) = x - \sqrt{3}$ $\frac{y}{2} - \frac{\pi}{4} = \tan^{-1}(x - \sqrt{3})$ $\frac{y}{2} = \frac{\pi}{4} + \tan^{-1}(x - \sqrt{3})$ $y = \frac{\pi}{2} + 2\tan^{-1}(x - \sqrt{3})$ Multiplying by 2 isolates the dependent variable $y$.

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Final Answer

Through variable separation and the $t = \tan(y/2)$ substitution, evaluated at $(0, -\pi/6)$: $\boxed{y = \frac{\pi}{2} + 2 \tan^{-1}(x - \sqrt{3})}$

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Common Mistakes

• Substitution Error: Forgetting to change $dy$ to $\frac{2}{1+t^2} dt$ when using the Weierstrass substitution.
• Arithmetic with $\tan(\pi/12)$: Students often struggle to calculate the exact value of $\tan(15^\circ)$, which is necessary to find the exact value of the constant $C$.
• Algebraic Rearrangement: Failing to use the tangent subtraction identity to transform the rational expression of $t$ back into a single $\tan^{-1}$ function.