Author Topic: QUIZ2 TUT0401  (Read 292 times)

baixiaox

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QUIZ2 TUT0401
« on: October 04, 2019, 02:00:07 PM »
Solve \begin{align*}
    (2xy^2 + 2y) + (2x^2y + 2x)y' &= 0\\
    \implies (2xy^2 + 2y)dx + (2x^2y + 2x)dy &= 0
\end{align*}

Let $M = 2xy^2 + 2y$ and $N = 2x^2y + 2x$
\begin{equation}
    M_y = \frac{d M}{dy} = 4xy + 2x\quad\quad\quad N_x = \frac{d N}{dy} = 4xy + 2x
\end{equation}

Since $M_y = N_x$,  the equation is exact so $\exists \psi(x, y)\quad s.t \quad \frac{\partial \psi}{\partial x} = M$ and $\frac{\partial \psi}{\partial y} = N$
Therefore
\begin{align*}
    \psi(x, y) &= \int M dx \\
    &= \int 2x^2y + 2y dx\\
    &= x^2y^2 + 2xy + h(y)
\end{align*}
Since
\begin{align*}
    \frac{\partial \psi}{\psi y} &= 2x^2y + 2x +h'(y)\\
    \implies h'(y) &= 0\\
    \implies h(y) &= C
\end{align*}

Therefore, the solution is $x^2y^2 + 2xy = C$