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\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

February 24, 2003}
\end{center}

\begin{enumerate}


\item  Using Euler's method, with step size $0.5$, estimate $y(3)$ for the initial value problem:
	\[
	\frac{dy}{dx} = y-xy \qquad y(2) = 4
	\]
	\\\sol
	\[
	\begin{array}{|r|r|r|}
	\hline
	n & x_n & y_n \\ \hline
	\hline
	0 & 2.0 & 4 \\ \hline
	1 & 2.5 & 2.0  \\ \hline
	2 & 3.0 & 0.5 \\ \hline
	\end{array}
	\]

\item   Consider the differential equation:
	\[
	\frac{dy}{dx} = \frac{x^7-12\sin^5x}{x^4+1}
	\]
	Is the following function a solution to this differential equation?  Why or why not?
	\[
	F(x) = \int_0^x \frac{t^7-12\sin^5t}{t^4+1} \; dt
	\]
	\\\sol
	Yes, $F(x)$ is a solution.  This is a review of the Fundamental Theorem of Calculus (part 1)
	-- do you remember how to take the derivative of an integral function?


\newpage 

\lecture

\item  Determine if the following differential equations are separable.
	After making this determination for all the equations, solve the separable equations.
	\begin{enumerate}
	\item  $\frac{dy}{dx} = 2x\sqrt{y}$
		\\\sol
		Separate: $y^{-1/2}\;dy = 2x \; dx$, then integrate to get:
		$2\sqrt{y} = x^2+C$.
		If you want, you can solve for $y$ to get $y=\frac{1}{4}(x^2+C)^2$

	\item  $\frac{dy}{dx} - 2y = 3e^{2x}$
		\\\sol Not separable.

	\item  $x\frac{dy}{dx} +3y = 2x^5 \quad y(2)=1$
		\\\sol Not separable.

	\item  $\frac{dy}{dx} = \frac{1}{4y^3} \quad y(0)=1$
		\\\sol
		Separate: $4y^3\;dy = dx$ and then integrate:
		$y^4=x+C$.
		Use the initial condition to get $C=1$ and the solution: $y^4=x+1$.

	\item  $\frac{dy}{dx} = 3x^2y^2-y^2 \quad y(0)=1$
		\\\sol First factor: $\frac{dy}{dx} = (3x^2-1)y^2$.  Now separate: 
		$y^{-2}\;dy = (3x^2-1)\; dx$.
		Integrate:
		$-\frac{1}{y} = x^3-x +C$.
		Solve for $y$ to get:
		$y = \frac{-1}{x^3-x+C}$.
		Use the initial condition to get $C=-1$ and the solution is $y=\frac{-1}{x^3-x-1}$.

	\item  $\frac{dy}{dx} + 2xy = x$
		\\\sol Rearrange to get $\frac{dy}{dx} = x(1-2y)$ and then separate:'
		$\frac{dy}{1-2y} = x \; dx$.
		Integrate to get $-\frac{1}{2}\ln|1-2y| = \frac{x^2}{2} +C$.
		You can rearrange to get either $\ln|1-2y| = -x^2 -2C$ or 
		$y=Ae^{-x^2}+\frac{1}{2}$ (where $A=-\frac{1}{2}e^{-2C}$).
	\end{enumerate}


\item  Find a family of curves orthogonal to the family $y=\frac{k}{x}$.
	\\\sol
	First take the derivative: $\frac{dy}{dx}=-\frac{k}{x^2}$.
	We now solve for $k$ in the original family to get $k=xy$ and substitute this into the DE.
	This gives $\frac{dy}{dx} = -\frac{y}{x}$.
	This is the differential equation representing the \emph{given} family.
	The differential equation representing the orthogonal family will be given by the 
	negative reciprocal of this one.
	In other words, we need to solve $\frac{dy}{dx}=\frac{x}{y}$, which has the solution:
	\[
	y^2-x^2=C
	\]
	(you might have $y^2-x^2=2C$).
	Notice that these are two families of orthogonal hyperbolas.

\end{enumerate}



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