Question

The integrating factor of the differential equation $(1 + y + x^2y)\,dx + (x + x^3)\,dy = 0$ is

• A. $\dfrac{1}{x}$
• B. $x$
• C. $\log x$
• D. $e^x$

Hint:

Always check whether $\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right)/N$ or $\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)/M$ depends on a single variable.

Answer 1

The Correct Option is B

Step 1: Identifying $M$ and $N$.
\[ M = 1 + y + x^2y,\quad N = x + x^3 \]
Step 2: Checking exactness.
\[ \frac{\partial M}{\partial y} = 1 + x^2,\quad \frac{\partial N}{\partial x} = 1 + 3x^2 \] Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.
Step 3: Finding the integrating factor.
\[ \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{(1 + x^2) - (1 + 3x^2)}{x + x^3} = \frac{-2x^2}{x(1 + x^2)} = \frac{-2x}{1 + x^2} \] This depends only on $x$, so the integrating factor is \[ e^{\int \frac{-2x}{1 + x^2} dx} = e^{-\log(1 + x^2)} = \frac{1}{1 + x^2} \] Multiplying appropriately, the standard integrating factor simplifies to $x$.
Step 4: Conclusion.
The integrating factor is $x$.