### Author Topic: Quiz 2 TUT0401  (Read 438 times)

#### NANAC

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##### Quiz 2 TUT0401
« on: October 04, 2019, 02:00:33 PM »
Here is the quiz question and solution

\left(3 x^{2} y+2 x y+y^{3}\right)+\left(x^{2}+y^{2}\right) y^{\prime}=0

Set

M(x, y)=3 x^{2} y+2 x y+y^{3} \quad N(x, y)=x^{2}+y^{2}

Then
\begin{eqnarray}
M_y(x, y)&=&\frac{\partial}{\partial y}\left(3 x^{2} y+2 x y+y^{3}\right)=3 x^{2}+2 x+3 y^{2} \\
N_x(x, y)&=&\frac{\partial}{\partial x}\left(x^{2}+y^{2}\right)=2 x
\end{eqnarray}
Since $M_y$ is not equal to $N_x$, the given differential equation is not exact.
\begin{eqnarray}
R&=&\frac{M y-N x}{N}=\frac{3 x^{2}+2 x+3 y^{2}-2 x}{x^{2}+y^{2}}=\frac{3\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)}=3 \\
\mu(x, y)&=&e^{\int R d x}=e^{\int 3 d x}=e^{3 x}
\end{eqnarray}
The given integrating factor is  $\mu(x, y)=e^{3 x}$, multiplying both sides of the given equation with

e^{3 x}\left(3 x^{2} y+2 x y+y^{3}\right) d x+e^{3 x}\left(x^{2}+y^{2}\right) d y=0

We have
\begin{eqnarray}
M(x, y)&=&3 x^{2} y e^{3 x}+2 x y e^{3 x}+y^{3} e^{3 x}\\
N(x, y)&=&e^{3 x} x^{2}+e^{3 x} y^{2}
\end{eqnarray}
Thus
\begin{eqnarray}
\begin{aligned}
M_y(x, y)&=\frac{\partial}{\partial y}\left(3 x^{2} y \cdot e^{3 x}+2 x y e^{3 x}+y^{3}  e^{3 x}\right)=2 x e^{3 x}+3 x^{2} e^{3 x}+3 y^{2}  e^{3 x},\\
N_{x}(x, y) &=\frac{\partial}{\partial x}\left(e^{3 x} x^{2}+e^{3 x} y^{2}\right) \\
&=2 x e^{3 x}+3 x^{2} e^{3 x}+3 y^{2}  e^{3 x}.
\end{aligned}
\end{eqnarray}
Since $M_y=N_x$, the differential equation is exact now.
There exist
\begin{eqnarray}
\begin{aligned}
\psi(x, y) \text { s.t } \psi(y) &=x^{2} e^{3 x}+y^{2} e^{3 x} \\ \psi(x, y) &=\int x^{2} e^{3 x}+y^{2} e^{3 x} d y \\ &=x^{2} y e^{3 x}+\frac{1}{3} y^{3} e^{3 x}+h(x) \end{aligned}
\end{eqnarray}

\psi_x=2x e^{3x} y+3x^2 e^{3x}y+y^3 e^{3x}+h'(x)=3x^2y e^{3x}+2y e^{3x}x+y^3e^{3x}

gives $h'(x)=0$, which implies $h(x)=C$.
Substitute $h(x)=C$  in equation
\begin{eqnarray}
\begin{aligned} \psi(x, y)&=x^{2} y e^{3 x}+\frac{1}{3} y^{3} e^{3 x}+h(x) \\
\psi(x, y) &=x^{2}y e^{3 x} +\frac{1}{3} y^{3} e^{3 x}+C \\ &=x^{2} ye^{3 x} +\frac{1}{3} y^{3} e^{3 x} \end{aligned}
\end{eqnarray}
Then the solution of differential equation is
\begin{eqnarray}
\begin{array}{c}{}  {\qquad \begin{array}{r}{x^{2} e^{3 x} y+\frac{1}{3} y^{3} e^{3 x}=k} \\ {3 x^{2} e^{3 x} y+y^{3} e^{3 x}=3 k} \\ \end{array}}\end{array}
\end{eqnarray}
We have

{\left(3 x^{2} y+y^{3}\right) e^{3 x}=C \quad (3 k=C)}

Hence, the required solution of the differential equation is
\begin{eqnarray}
\left(3x^{2} y+y^{3}\right) e^{3 x}=C
\end{eqnarray}