Question

The solution(s) of the differential equation \( \frac{dy}{dx} = (\sin 2x) y^{1/3}\) satisfying  \(y(0) = 0\) is(are)
 

• A. \( y(x) = 0 \)
• B. \( y(x) = -\sqrt{\dfrac{8}{27}} \sin^3 x \)
• C. \( y(x) = \sqrt{\dfrac{8}{27}} \sin^3 x \)
• D. \( y(x) = \sqrt{\dfrac{8}{27}} \cos^3 x \)

Hint:

When solving differential equations with power terms like \( y^{1/3} \), use separation of variables and apply initial conditions to find the specific solution.

Answer 1

The Correct Option is A, B, C

Step 1: Separate variables

This is a separable ODE. Rearranging: $$\frac{dy}{y^{1/3}} = \sin 2x , dx$$

$$y^{-1/3} dy = \sin 2x , dx$$

Step 2: Integrate both sides

$$\int y^{-1/3} dy = \int \sin 2x , dx$$

Left side: $$\int y^{-1/3} dy = \frac{y^{2/3}}{2/3} = \frac{3}{2} y^{2/3}$$

Right side: $$\int \sin 2x , dx = -\frac{1}{2}\cos 2x + C$$

Therefore: $$\frac{3}{2} y^{2/3} = -\frac{1}{2}\cos 2x + C$$

Step 3: Apply initial condition $y(0) = 0$

$$\frac{3}{2} \cdot 0^{2/3} = -\frac{1}{2}\cos(0) + C$$

$$0 = -\frac{1}{2}(1) + C$$

$$C = \frac{1}{2}$$

Step 4: Solve for $y$

$$\frac{3}{2} y^{2/3} = -\frac{1}{2}\cos 2x + \frac{1}{2}$$

$$\frac{3}{2} y^{2/3} = \frac{1}{2}(1 - \cos 2x)$$

$$y^{2/3} = \frac{1}{3}(1 - \cos 2x)$$

Using the identity $1 - \cos 2x = 2\sin^2 x$:

$$y^{2/3} = \frac{2\sin^2 x}{3}$$

$$y = \left(\frac{2\sin^2 x}{3}\right)^{3/2} = \frac{2^{3/2}}{3^{3/2}} (\sin^2 x)^{3/2}$$

$$y = \frac{2\sqrt{2}}{3\sqrt{3}} \sin^3 x = \frac{2\sqrt{2}}{3\sqrt{3}} \sin^3 x$$

Simplifying: $$y = \frac{2\sqrt{2}}{3\sqrt{3}} \sin^3 x = \frac{2\sqrt{2} \cdot \sqrt{3}}{3 \cdot 3} \sin^3 x = \frac{2\sqrt{6}}{9} \sin^3 x$$

$y^{2/3} = \frac{2\sin^2 x}{3}$

$$y = \pm \left(\frac{2\sin^2 x}{3}\right)^{3/2} = \pm \frac{2^{3/2} \sin^3 x}{3^{3/2}} = \pm \frac{2\sqrt{2}}{3\sqrt{3}} |\sin x|^3 \cdot \text{sgn}(\sin x)$$

For continuous solutions passing through $(0,0)$: $$y = \pm \frac{2\sqrt{2}}{3\sqrt{3}} \sin^3 x = \pm \sqrt{\frac{8}{27}} \sin^3 x$$

Step 5: Consider the trivial solution

Note that $y(x) = 0$ also satisfies the differential equation since: $$\frac{dy}{dx} = 0 = \sin 2x \cdot 0^{1/3}$$

And it satisfies $y(0) = 0$.

Analysis of Options

(A) $y(x) = 0$:

This is the trivial solution and satisfies both the ODE and initial condition. 

(B) $y(x) = -\sqrt{\frac{8}{27}} \sin^3 x$:

This is one of the non-trivial solutions. 

(C) $y(x) = \sqrt{\frac{8}{27}} \sin^3 x$:

This is another non-trivial solution. 

(D) $y(x) = \sqrt{\frac{8}{27}} \cos^3 x$:

Check: $y(0) = \sqrt{\frac{8}{27}} \cos^3(0) = \sqrt{\frac{8}{27}} \neq 0$ 

Answer: (A), (B), and (C)