Solve the differential equation 4x^3y + x^4y' = sin^3 x.

Steps to Solve

1. Rearranging the differential equation We start with the equation: $$ 4x^3y + x^4y' = \sin^3 x $$ Rearranging it gives: $$ x^4y' = \sin^3 x - 4x^3y $$

2. Separating variables Next, we will express the equation in a form suitable for separating variables: $$ y' = \frac{\sin^3 x - 4x^3y}{x^4} $$ This implies: $$ \frac{dy}{dx} = \frac{\sin^3 x}{x^4} - \frac{4y}{x} $$

3. Finding an integrating factor To solve this linear first-order differential equation, we can find the integrating factor: The equation can be rewritten as: $$ y' + \frac{4}{x}y = \frac{\sin^3 x}{x^4} $$ The integrating factor $\mu(x)$ is given by: $$ \mu(x) = e^{\int \frac{4}{x} dx} = e^{4 \ln |x|} = |x|^4 $$

4. Multiplying the whole equation by the integrating factor Now, we multiply the entire equation by the integrating factor: $$ |x|^4 y' + 4|x|^3 y = |x|^4 \cdot \frac{\sin^3 x}{x^4} $$ This simplifies to: $$ |x|^4 y' + 4|x|^3 y = \sin^3 x $$

5. Expressing the left side as a derivative The left side can be expressed as the derivative of a product: $$ \frac{d}{dx}(|x|^4 y) = \sin^3 x $$

6. Integrating both sides Now we integrate both sides: $$ \int \frac{d}{dx}(|x|^4 y) , dx = \int \sin^3 x , dx $$ This gives: $$ |x|^4 y = -\frac{1}{3} \cos(3x) + C $$

7. Solving for y Finally, we isolate (y): $$ y = \frac{-\frac{1}{3} \cos(3x) + C}{|x|^4} $$