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\noindent{\bf Math 3226
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Laboratory  1C}\\
\begin{center}
{\Large\bf A Substitution Technique}
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\noindent{\em You will explore a substitution technique which transforms certain differential equations into  separable differential equations, which can then be solved by the method discussed in class.}\medskip

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\noindent{\bf 1 Warm-up Exercise.} Find the general solution of the following differential equations. Express the solutions in explicit form, whenever possible.
\begin{enumerate}
\item \ilf{\frac{dy}{dt}=\frac{3t^2 y}{1+t^3}}
\item \ilf{\frac{dy}{dt}=\frac{5y}{t (y+2)}}
\item \ilf{\frac{dy}{dt}=\sqrt{1-y^2}}
\item \ilf{\frac{dy}{dt}=\frac{1}{y\sqrt{1-y^2}}}
\end{enumerate}\medskip

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\noindent{\em   For $t>0$ consider the differential equation
\begin{equation}\frac{dy}{dt}=\frac{y^2+ty}{t^2}\end{equation} 
This differential equation is neither separable nor linear.
We will now perform a substitution: we will replace $y(t)$ in the equation by 
\[z(t)=\frac{y(t)}{t},\]
or equivalently \[y(t)=t z(t).\]
We will also have to replace \ilf{\frac{dy}{dt}}  by an expression containing $z(t)$ and $z'(t)$, using the product rule:
\[\frac{dy}{dt}=z+t \frac{dz}{dt}.\]
Performing the substitution on both sides of the differential equation {\rm (1)} reads as:
\begin{equation}z+t \frac{dz}{dt}=z^2+z.\end{equation}
Simplifying, we obtain the {\bf separable} differential equation:
\begin{equation}t \frac{dz}{dt}=z^2.\end{equation}\medskip}

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\noindent{\bf 2 } Show that equation (3) has the solutions
\ilf{z(t)=\frac{1}{C-\ln t}}\medskip

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\noindent{\em Finally we re-substitute: The solutions to the original equation {\rm (1)} are given by
\[y(t)=t z(t)=\frac{t}{C-\ln t}.\]\medskip

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\noindent What is special about the differential equation {\rm (1)}? It can be written in the form \ilf{\frac{dy}{dt}=F\left(\frac{y}{t}\right)}, i.e. the right hand side depends only on the {\bf quotient of $y$ and $t$} and not on $y$ and $t$ independently.}\medskip

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\noindent{\bf 3 } What is this function $F$ for the differential equation (1)?\medskip

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\noindent {\em Differential equations of the form \ilf{\frac{dy}{dt}=F\left(\frac{y}{t}\right)} are called {\bf homogeneous}.}\medskip

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\noindent{\bf 4 } Show that the substitution \ilf{z(t)=\frac{y(t)}{t}}  transforms a  homogeneous differential equation into a separable differential equation (in $z(t)$)!\medskip

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\noindent{\bf 5 } Solve the following differential equations. Express the solutions in explicit form if possible.
\begin{enumerate}
\item \ilf{\frac{dy}{dt}=\frac{t^2+ty+y^2}{t^2} }
\item \ilf{\frac{dy}{dt}=\frac{y-t}{y+t}}
\item \ilf{\frac{dy}{dt}=-\frac{t^2+y^2}{ty} }
\item \ilf{\frac{dy}{dt}=\frac{y}{t}+\sin\left(\frac{y}{t}\right) }
\end{enumerate}\medskip

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\noindent{\bf 6 } Investigate the slope fields of homogeneous differential equations: How can you tell from the slope field that a given differential equation is homogeneous? {\bf Hint}: Find the lines on which the slope is constant. \medskip

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\noindent{\bf 7 } Investigate the slope field of the differential equation
\begin{equation}\frac{dy}{dt}=\frac{y-t+3}{y+t-1}.\end{equation}
Using your observations in {\bf 6}, explain why the differential equation is not homogeneous. Compare  the  slope field  to  the  slope field of the homogeneous equation
\begin{equation}\frac{dy}{dt}=\frac{y-t}{y+t}.\end{equation}
Find a substitution method to transform the differential equation (4) into the homogeneous differential equation (5), then solve  differential equation (4).
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\flushright\tiny\copyright\  H. Knaust. \today .
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