Solve the differential equation. ,

Question

Solve the differential equation. ​  ,  ​

Quizplus · Accepted Answer

To solve this differential equation, we can first separate the variables:

$$\frac{dy}{dx} = \frac{y^2(x^2 + 1)}{x^4}$$

$$\frac{1}{y^2}dy = \frac{x^2 + 1}{x^4} dx$$

Integrating both sides:

$$-\frac{1}{y} = -\frac{1}{x} - \frac{1}{3x^3} + C$$

Multiplying both sides by $-1$ and solving for $y$:

$$y = \frac{3x^3}{Cx^4 - x - 1}$$

Now, let's use the initial condition $y(1) = 2$ to solve for $C$:

$$2 = \frac{3C}{C - 2}$$

$$2C - 4 = 3C$$

$$C = -4$$

Substituting this value back into our expression for $y$:

$$y = \frac{3x^3}{4x^4 + x + 1}$$

Therefore, the correct choice is B.

Solve the Differential Equation