Question

$(x^2 + 1)2dy + (y(2x^3 + x) – 2)dx = 0, y(0) = 0,$ then y(2) is equal to

• A. $\frac{2}5 tan^{-1}2 $
• B. $\frac{3}5 tan^{-1}2 $
• C. $\frac{2}5 tan^{-1}3$
• D. $\frac{3}5 tan^{-1}3 $

Answer 1

The Correct Option is A

To solve the differential equation given by (x^2 + 1)2dy + (y(2x^3 + x) – 2)dx = 0 with the initial condition y(0) = 0, we need to follow these steps:

1. Rewrite the differential equation in the standard form:

   The equation is originally (x^2 + 1)2dy + (y(2x^3 + x) – 2)dx = 0. This can be rewritten as:

   2(x^2 + 1) \frac{dy}{dx} = 2 - y(2x^3 + x)

2. Separate the variables:

   To separate variables, rearrange the equation:

   2 \frac{dy}{y} = \frac{(2 - (2x^3 + x) dx}{x^2 + 1}

3. Integrate both sides:

   Integrating both sides gives:

   \int \frac{2}{y} \, dy = \int \frac{2 - (2x^3 + x)}{x^2 + 1} \, dx

   Resulting in:

   2 \ln|y| = \int \frac{2}{x^2+1} \, dx - \int \frac{2x^3}{x^2+1} \, dx - \int \frac{x}{x^2+1} \, dx

4. Compute the integrals:

   • \int \frac{2}{x^2+1} \, dx = 2 \tan^{-1} x

   • For the integral \int \frac{2x^3}{x^2+1} \, dx, use substitution:

     If we set u = x^2 + 1, then du = 2x \, dx. The integral becomes:

     \int x^2 \, du = \frac{x^3}{3} + C
   • \int \frac{x}{x^2+1} \, dx = \frac{1}{2} \ln|x^2 + 1|

5. Substitute back and solve for integration constant using the initial condition:

   2 \ln|y| = C + 2\tan^{-1} x - \frac{x^3}{3} - \frac{1}{2} \ln|x^2+1|

   Use the initial condition y(0) = 0 to find C.

6. Find y(2):

   Substitute x = 2 into the solved equation to find y(2).

Therefore, y(2) = \frac{2}{5} \tan^{-1} 2, which matches the correct answer from the options provided.