Question

The general solution of the differential equation $x^2(1+y^3)dx = y^2(1+x^3)dy$ is

• A. $(1+x^2)(1+y^2) = C$
• B. $1+x^3 = C(1+y^3)$
• C. $(x+y)(1+x^2+x^3) = C$
• D. $x(1+y^2) = Cy(1+x^2)$

Answer 1

Answer: (B)

1+x^3 = C(1+y^3)

Answer 2

The given differential equation is $x^2(1+y^3)dx = y^2(1+x^3)dy$. This is a first-order differential equation. We can separate the variables $x$ and $y$.

Divide both sides by $(1+x^3)(1+y^3)$, assuming $1+x^3 \neq 0$ and $1+y^3 \neq 0$:

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

$\frac{x^2}{1+x^3}dx = \frac{y^2}{1+y^3}dy$

Now, integrate both sides:

$\int \frac{x^2}{1+x^3}dx = \int \frac{y^2}{1+y^3}dy$

For the integral on the left side, let $u = 1+x^3$. Then $du = 3x^2 dx$. So $x^2 dx = \frac{1}{3}du$.

$\int \frac{x^2}{1+x^3}dx = \int \frac{1}{u} \left(\frac{1}{3}du\right) = \frac{1}{3}\int \frac{1}{u}du = \frac{1}{3}\ln|u| + C_1 = \frac{1}{3}\ln|1+x^3| + C_1$.

For the integral on the right side, let $v = 1+y^3$. Then $dv = 3y^2 dy$. So $y^2 dy = \frac{1}{3}dv$.

$\int \frac{y^2}{1+y^3}dy = \int \frac{1}{v} \left(\frac{1}{3}dv\right) = \frac{1}{3}\int \frac{1}{v}dv = \frac{1}{3}\ln|v| + C_2 = \frac{1}{3}\ln|1+y^3| + C_2$.

Equating the results of the integration:

$\frac{1}{3}\ln|1+x^3| + C_1 = \frac{1}{3}\ln|1+y^3| + C_2$

$\frac{1}{3}\ln|1+x^3| - \frac{1}{3}\ln|1+y^3| = C_2 - C_1$

$\frac{1}{3}(\ln|1+x^3| - \ln|1+y^3|) = C_3$, where $C_3 = C_2 - C_1$ is an arbitrary constant.

$\frac{1}{3}\ln\left|\frac{1+x^3}{1+y^3}\right| = C_3$

$\ln\left|\frac{1+x^3}{1+y^3}\right| = 3C_3$

$\left|\frac{1+x^3}{1+y^3}\right| = e^{3C_3}$

$\frac{1+x^3}{1+y^3} = \pm e^{3C_3}$

Let $C = \pm e^{3C_3}$. Since $C_3$ is an arbitrary constant, $e^{3C_3}$ is an arbitrary positive constant. Thus, $C$ is an arbitrary non-zero constant.

$1+x^3 = C(1+y^3)$, where $C \neq 0$.