Differential Equations
Exact ODE; value at a point
MJMT_Full_Test_07
Grade 12
Question:
If $(2xy-y^2-y)dx=(2xy+x-x^2)dy$ and $y(1)=1$, then the value of $12|y(-1)|$ is
Step-by-Step Solution
Key Concept: Rearrange to exact form. $M=2xy-y^2-y$, $N=-(2xy+x-x^2)$. Check exactness, find integrating factor if needed. Solve: $x^2y-xy^2-xy=c$. With $y(1)=1$: $c=-1$. Find $y(-1)$.
$12|y(-1)|=12$.
Correct Answer: 1