### Author Topic: TUT0303 Quiz2  (Read 917 times)

#### Tiantian Yu

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##### TUT0303 Quiz2
« on: October 04, 2019, 05:57:49 PM »
find the value of b for which the given equation is exact, and then solve it using that value of b.
\begin{equation*}
(xy^2+bx^2y)+(x+y)x^2y'=0
\end{equation*}
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Solution: By calculating My and Nx , we find that
\begin{equation*} M_y(x,y)= \frac{d}{dy} (xy^2+bx^2y) = 2xy+bx^2 \end{equation*} \begin{equation*} N_x(x,y)= \frac{d}{dx} ((x+y)x^2) =  \frac{d}{dx} (x^3+yx^2)= 3x^2+2xy \end{equation*}
since we know that the given equation is exact, and it is exact when My=Nx
\begin{equation*}2xy+bx^2=3x^2+2xy\end{equation*}
thus we get b=3

there is a ψ(x, y) such that
\begin{equation*}ψ_x(x,y)=xy^2+3x^2y \end{equation*} \begin{equation*}ψ_y(x,y)=(x+y)x^2=x^3+yx^2\end{equation*}
Integrating the first of these equations with respect to x, we obtain \begin{equation*}ψ (x , y )=\int(xy^2+3x^2y) dx=\frac{1}{2} x^2y^2+x^3y+h(y) \end{equation*}
Then computing ψy from the equation above and setting ψy = N gives
\begin{equation*}ψ_y(x,y) =\frac{d}{dy} (\frac{1}{2} x^2y^2+x^3y+h(y))=x^2y+x^3+h'(y)=x^3+yx^2\end{equation*}
Thus h′(y)=0, and h(y)=C
Substituting for h(y) in ψ(x, y) gives \begin{equation*}ψ(x,y) =\frac{1}{2} x^2y^2+x^3y+C \end{equation*}
Hence the solution is given implicitly by \begin{equation*}C=\frac{1}{2} x^2y^2+x^3y=x^2y^2+2x^3y\end{equation*}