Question

A curve passing through origin and satisfying the differential equation $\frac{dy}{dx}=\left[\frac{1}{(1+x^3)}\right]-\left[\frac{3x^2}{(1+x^3)}\right]y,$ where $x>-1$ is:

• A. $y=\frac{x}{(1-x^3)}$
• B. $y=\frac{x}{(1+x)}$
• C. $y=\frac{x}{(1+x^2)}$
• D. $y=\frac{x}{(1+x^3)}$

Answer 1

The correct option is D $y=\frac{x}{(1+x^3)}$

The above mentioned equation can be rewritten as $\frac{dy}{dx}+\left[\frac{3x^2}{(1+x^3)}\right]y=\frac{1}{(1+x^3)}$
Comparing it with $\frac{dy}{dx}+Py=Q,$ we get
$P=\frac{3x^2}{1+x^3}$ and $Q=\frac{1}{1+x^3}$

Now,$\text{I.F}=e^{\int \frac{3x^2}{1+x^3}dx}=e^{\ln |1+x^3|}$
$\Rightarrow \text{I.F.}=1+x^3$
So, the solution can be obtained as:$y\times (1+x^3)=\int \left[\frac{1}{(1+x^3)}\right]\times (1+x^3)dx$
$\Rightarrow y\times (1+x^3)=x+C$
Since the curve passes through origin, we have $C=0$
Hence we have $y=\frac{x}{(1+x^3)}$