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\begin{document}

\begin{center}
\textbf{Math 2233.21570}\\[0pt]
\end{center}

1. 

(a) $\displaystyle\frac{d^{2}x}{dt^{2}}+tx=\sin (x):$ 2nd order, linear, ODE

(b) $\displaystyle\frac{\partial ^{3}\psi }{\partial ^{3}x}-x^{2}\frac{%
\partial \psi }{\partial y}=\psi ^{2}:$ 3rd order, nonlinear, PDE

(c) $\displaystyle\frac{d^{3}x}{dt^{3}}+t^{2}\frac{dx}{dt}+x=0:$ 3rd order,
nonlinear, ODE

(d) $\displaystyle x^{2}y^{\prime }+y=e^{x}\sin \left( x\right) :$ 1st
order, linear, ODE

(e) $\displaystyle\frac{\partial ^{2}\phi }{\partial x^{2}}+\frac{\partial
\phi }{\partial y}=\left( x+y\right) ^{2}:$ 2nd order, linear, PDE

2. 
\begin{eqnarray*}
F\left( x,y\right)  &=&xy \\
y_{1} &=&2.2 \\
y_{2} &=&2.442 \\
y_{3} &=&2.735
\end{eqnarray*}

3. $\quad 3x^{2}-e^{2y}y^{\prime }=x$ 
\begin{eqnarray*}
M\left( x\right)  &=&3x^{2}-x\quad ,\quad N\left( y\right) =-e^{2y} \\
x^{3}-\frac{1}{2}x^{2}-\frac{1}{2}e^{2y} &=&C \\
y\left( x\right)  &=&\frac{1}{2}\ln \left\vert 2x^{3}-x^{2}-2C\right\vert 
\end{eqnarray*}%
\ 

4. $\qquad xy^{\prime }-2y=x^{2}\qquad ,\qquad y(1)=2$ 
\begin{eqnarray*}
y^{\prime }-\frac{2}{x}y &=&x\quad \Rightarrow \quad p\left( x\right) =-%
\frac{2}{x}\quad ,\quad g\left( x\right) =x \\
\mu \left( x\right)  &=&\exp \left[ \int -\frac{2}{x}dx\right] =x^{-2} \\
y\left( x)=\right) x^{2}\int \left( x^{-2}\right) \left( x\right) dx+Cx^{2}
&=&x^{2}\ln \left\vert x\right\vert +Cx^{2}
\end{eqnarray*}

5.$\qquad 3x^{2}+\left( 2y-2x\right) \frac{dy}{dx}=2y\quad ,\quad y\left(
1\right) =2$ 
\begin{eqnarray*}
M\left( x,y\right)  &=&3x^{2}-2y\qquad ,\qquad N\left( x,y\right)
=2y-2x\quad \Rightarrow \quad \frac{\partial M}{\partial y}=-2=\frac{%
\partial N}{\partial x} \\
\Phi \left( x,y\right)  &=&\int \left( 3x^{2}-2y\right) \partial
x+c_{1}\left( y\right) =x^{3}-2yx+c_{1}\left( y\right)  \\
\Phi \left( x,y\right)  &=&\int \left( 2y-2x\right) \partial y+c_{2}\left(
x\right) =y^{2}-2xy+c_{2}\left( x\right)  \\
\Phi \left( x,y\right)  &=&x^{3}-2xy+y^{2}=C \\
y\left( x\right)  &=&x+\sqrt{x^{2}-x^{3}+1}
\end{eqnarray*}

6. $x\frac{dy}{dx}=x+y$

\begin{eqnarray*}
\frac{dy}{dx} &=&1+\left( \frac{y}{x}\right)  \\
x\frac{dz}{dx}+z &=&z+1 \\
-\frac{1}{x}+\frac{dz}{dx} &=&0 \\
-\ln \left\vert x\right\vert +z &=&C\quad  \\
y &=&xz=x\ln \left\vert x\right\vert +Cx
\end{eqnarray*}

\end{document}
