Author Topic: TUT0401 Quiz2  (Read 276 times)

Yuying Chen

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TUT0401 Quiz2
« on: October 04, 2019, 02:00:12 PM »
$Question: (2xy^2+2y)+(2x^2y+2x)y^{\prime}=0\\$
$M=2xy^2+2y\qquad M_{y}=\frac{\partial}{\partial y}M=4xy+2\\$
$N=2x^2y+2x\qquad N_{x}=\frac{\partial}{\partial x}N=4xy+2\\$
$\text{Since $M_{y}=N_{x}$, the given differential equation is exact.}\\ $
$\text{$\exists \psi{(x,y)}$ such that $\psi_{x}=M=2xy^2+2y$}\\$
$\qquad\quad\psi{(x,y)}=\int {(2xy^2+2y)dx}\\$
$\qquad\qquad\qquad =x^2y^2+2xy+h(y)\\$
$\qquad\quad\psi_{y}=2x^2y+2x+h^{\prime}(y)\\$
$\text{set $\psi_{y}=N$}\\$
$\text{Therefore,}\\$
$\qquad\quad h^{\prime}(y)=0\\$
$\qquad\quad h(y)=0\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=x^2y^2+2xy=C\\$
$\text{Hence the solutions of the given differential equation are implicitly by}\\$
$\qquad\quad x^2y^2+2xy=C$