Differential Equations
Exact equation via substitution; initial value problem
Grade Class 12
Question:
Let $y=y(x)$ satisfy the differential equation $\left(2xy + x^2y + \dfrac{y^3}{3}\right)dx + \left(x^2+y^2\right)dy=0$. If $y(1)=1$ and $(y(0))^3=ke$, $k\in\mathbb{N}$, then $k$ is
Step-by-Step Solution
Key Concept: Identify exact groupings: $(2xydx + x^2dy) + (x^2ydx) + (\frac{y^3}{3}dx + y^2dy)=0$. Let $t=x^2y$ and $u=y^3/3$.
$x^2y+y^3/3 = \frac{4}{3}e^{-(x-1)}$. At $x=0$: $y^3/3=\frac{4e}{3} \Rightarrow y^3=4e \Rightarrow k=4$.
Correct Answer: 2