Differential Equations
Exact ODE
MMTS_Full_Test_11
Grade 12

Question:

The solution of $(1+y+x^2y)dx+(x+x^3)dy=0$ is
$y=\dfrac{\tan^{-1}x+c}{x}$
$xy=\tan^{-1}x+c$
$y+\tan^{-1}x=cx$
$xy+\tan^{-1}x=c$

Step-by-Step Solution

Key Concept: Check for exact ODE or integrating factor; note $x+x^3=x(1+x^2)$
$dx+(1+x^2)(y\,dx+x\,dy)=0\Rightarrow dx+(1+x^2)d(xy)=0$. Let $u=xy$: $\int\frac{dx}{1+x^2}+u=c... $ Actually $\frac{dx}{1}+(1+x^2)du/(1)=0$... $\frac{du}{dx}=-(1+x^2)^{-1}/(1)$. $u=-\tan^{-1}x+c$. $xy+\tan^{-1}x=c$.
Correct Answer: 1

Master Differential Equations with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free