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 given differential equation and find \( y(2) \) when \( y(0) = 0 \), we first rewrite the given equation:

(x^2 + 1)2dy + (y(2x^3 + x) - 2)dx = 0

This can be expressed as:

2(x^2 + 1) dy = (2 - y(2x^3 + x)) dx

Rewriting, we get:

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

The equation can be solved by separating variables:

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

Next, we integrate both sides:

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

Solving the integral on the right side:

\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x) + C

Now for the integral on the left:

Using substitution, let u = 2 - y(2x^3 + x) \Rightarrow \frac{dy}{u} = -\frac{2}{x^2 + 1} . This simplifies into a standard form:

\int 2 du = \int \frac{2}{x^2 + 1} dx \Rightarrow u = -\ln | u | + C_1

Equating both sides:

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

Apply the initial condition \( y(0) = 0 \):

- \ln | 2 - 0(2 \cdot 0^3 + 0) | = 0 \Rightarrow C = 0

So we have:

-\ln | 2 - y(2x^3 + x) | = \tan^{-1}(x)

Inverting and substituting \( x = 2 \), compute \( y(2) \):

y(2) = \frac{2}{5} \tan^{-1}(2)

Hence, the correct answer is:
\frac{2}{5}\tan^{-1}(2)