Question

Solve the given DE or IVP. First you need to determine what type of DE it is.

y'=sin^2 (3x-3y+1)

Answer 1

Answer on Question #82599 – Math – Differential Equations

Solve the given DE or IVP. First you need to determine what type of DE it is.

$$y' = \sin^2 (3x - 3y + 1)$$

Solution

The first-order nonlinear ODE

$$\begin{aligned} \text{let } z(x) = 3x - 3y(x) + 1 &\to \frac{dz(x)}{dx} = 3 - 3\frac{dy(x)}{dx} \to \frac{dy(x)}{dx} = 1 - \frac{1}{3}\frac{dz(x)}{dx} \\ \text{then } 1 - \frac{1}{3}\frac{dz(x)}{dx} = \left(\sin(z(x))\right)^2 &\quad \\ \frac{dz(x)}{\left(\cos(z(x))\right)^2} = 3dx &\to \int \frac{dz(x)}{\left(\cos(z(x))\right)^2} = \int 3dx \\ \tan(z(x)) = \text{Const} + 3x & \\ z(x) &= \tan^{-1}(C + 3x) + k \cdot \pi, \, k = 0, \pm 1, \dots, \pm \infty \\ \tan\alpha &= \frac{2 \tan\alpha / 2}{1 - (\tan\alpha / 2)^2} \to \tan\alpha / 2 = \frac{\sqrt{(\tan\alpha)^2 + 1} - 1}{\tan\alpha} \quad \text{and} \quad \tan\alpha / 2 = -\frac{\sqrt{(\tan\alpha)^2 + 1} + 1}{\tan\alpha} \\ \frac{\alpha}{2} &= \tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} - 1}{C + 3x}\right) \quad \text{or} \quad \frac{\alpha}{2} = -\tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} + 1}{C + 3x}\right) \\ z_1(x) &= 2 \tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} - 1}{C + 3x}\right), \quad z_2(x) = -2 \tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} + 1}{C + 3x}\right) \\ z(x) &= 3x - 3y(x) + 1 \quad \rightarrow \quad y_1(x) = \frac{1}{3} + x + \frac{2}{3} \tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} - 1}{C + 3x}\right), \quad y_2(x) \\ &= \frac{1}{3} + x - \frac{2}{3} \tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} + 1}{C + 3x}\right) \end{aligned}$$

Answer:

$$y_1(x) = \frac{1}{3} + x + \frac{2}{3} \tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} - 1}{C + 3x}\right), \quad y_2(x) = \frac{1}{3} + x - \frac{2}{3} \tan^{-1}\left(\frac{\sqrt{(C + 3x)^2 + 1} + 1}{C + 3x}\right)$$

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