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\noindent{\bf Math 2326
\hfill
Laboratory  1B}\\
\begin{center}
{\Large\bf A Substitution Technique}
\bigskip\end{center}
\noindent{\em You will explore a substitution technique which transforms certain differential equations into  linear 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:
\begin{enumerate}
\item $y'(t)-5y(t)=e^{2t}$
\item $y'(t)+2ty(t)=4t\phantom{e^t}$
\item \ilf{ty'(t)+2y(t)=e^{t^2}}
\item $(1+t^2)y'(t)-2ty(t)=2t(1+t^2)$ 
\end{enumerate}\medskip

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\noindent{\em  For $t>0$ consider the differential equation
\begin{equation}y'(t)+\frac{2}{t}y(t)=\frac{1}{t^2} (y(t))^3.\end{equation} 
This differential equation is not linear because of the $y(t)^3$ term.
We will now perform a substitution: we will replace $y(t)$ in the equation by 
\[z(t)=y(t)^{-2},\]
or equivalently \[y(t)=z(t)^{-1/2}.\]
We will also have to replace $y'(t)$  by an expression containing $z(t)$ and $z'(t)$, using the chain rule:
\[y'(t)=(-1/2) z(t)^{-3/2} z'(t).\]
Performing the substitution the differential equation {\rm (1)} reads as:
\begin{equation}(-1/2) z(t)^{-3/2} z'(t)+\frac{2}{t}z(t)^{-1/2}=\frac{1}{t^2}z(t)^{-3/2}.\end{equation}
Multiplying by the reciprocal of the term in front of $z'(t)$, we obtain the {\bf linear} differential equation:
\begin{equation}z'(t)-\frac{4}{t}z(t)=-\frac{2}{t^2}.\end{equation}\medskip}

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\noindent{\bf 2 } Show that equation (3) has the solutions
\ilf{z(t)=\frac{2}{5t}+C t^4}\medskip

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\noindent{\em Finally we re-substitute: The solutions to the original equation {\rm (1)} are given by
\[y(t)=z(t)^{-1/2}=\left(\frac{2}{5t}+C t^4\right)^{-1/2}.\]}\medskip

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\noindent{\bf 3 } Show that the substitution \ilf{z(t)=y(t)^{1-n}}  transforms a differential equation of the form
\begin{equation}y'(t)+p(t)y(t)=q(t)y(t)^n\end{equation} into a linear differential equation (in $z(t)$) for $n\not=0,1$!\medskip

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\noindent{\bf 4} Find the general solution of  equation (4) for the cases $n=0$ and $n=1$. \medskip

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\noindent{\em A differential equation of the type
\ilf{y'(t)+p(t)y(t)=q(t)y(t)^n} is called a {\bf Bernoulli Equation}. } \medskip

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\noindent{\bf 5 }Solve the following differential equations for the initial 
condition $y(2)=1$:
\begin{enumerate}
\item $y'(t)-y(t)=y^5(t)$
\item $y'(t)+\frac{2}{3}ty(t)=ty^4(t)$
\item \ilf{y'(t)-y(t)=\frac{e^t}{y(t)}}
\end{enumerate}\medskip

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\noindent{\bf 6} Solve the differential equations in {\bf 5} for the initial condition $y(2)=-1$. (You must slightly adapt the method for this initial condition! Why?)\medskip

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\noindent{\bf 7} The method of solving Bernoulli equations does not work when the initial condition is chosen so that $y=0$. Why? (This does not mean necessarily that a solution does not exist.)
What can you say about the solutions to the differential equations in {\bf 5} satisfying the initial condition $y(2)=0$?

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