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**MAT244--Lectures & Home Assignments / LEC 5101-corrections to Lecture of Sept. 10**

« **on:**September 11, 2019, 04:00:52 AM »

In two examples were made errors.

\begin{equation}

y'= -\frac{y}{x}+y^2.

\label{1}

\end{equation}

It is

\begin{equation}

y'= -\frac{y}{x}.

\label{2}

\end{equation}

It has a solution

\begin{equation}

y= Cx^{-1}

\label{3}

\end{equation}

with constant $C$ (do it by yourself!). Now consider (\ref{3}) with $C$ which is not a constant (variation!) and plug it into (\ref{1}).

\begin{gather*}

(Cx^{-1})' =-x^{-1}(Cx^{-1})+(Cx^{-1})^2\implies C'x^{-1}-Cx^{-2}=-Cx^{-2} + C^2x^{-2}\implies C' =C^2x^{-1}\\

\implies \frac{dC}{C^2}=\frac{dx}{x}\implies -\frac{1}{C}=\ln(x)+c

\end{gather*}

where $c=\mathsf{const}$. Then $C=-\dfrac{1}{\ln (x)+c}$ and plugging into (\ref{3}) we get

\begin{equation*}

\boxed{ y=-\frac{1}{x(\ln (x)+c)}.}

\end{equation*}

\begin{equation}

y'- \tan(x) y =\cos(x).

\label{4}

\end{equation}

We solve it by

\begin{equation}

\mu y' - \mu \tan (x) y = \mu \cos(x).

\label{5}

\end{equation}

\begin{equation*}

\mu'=-\mu \tan(x) \implies \frac{d\mu}{\mu}= \tan(x)\,dx \implies \ln (\mu) =-\int \tan(x)\,dx = -\int\frac{\sin(x)}{\cos(x)}\,dx = \ln (\cos(x)).

\end{equation*}

You

So $\mu =\cos(x)$. (\ref{5}) is now

\begin{gather*}

\cos(x)y'-\sin (x)y =\cos^2(x)\implies \bigl(\cos(x)y\bigr)'= \cos^2(x)\implies

\cos(x)y = \int \cos^2(x)\,dx = \int \frac{1+\cos(2x)}{2}\,dx =\\

\frac{x}{2}+\frac{\sin (2x)}{4}+C=

\frac{x}{2}+\frac{\sin (x)\cos(x)}{2}+C.

\end{gather*}

You

\begin{equation*}

\boxed{y= \frac{1}{2}(x+2C)\sec(x)+ \frac{1}{2}\sin (x).}

\end{equation*}

**A.**Consider equation\begin{equation}

y'= -\frac{y}{x}+y^2.

\label{1}

\end{equation}

It is

*Bernoulli equation*. We solve it by the*method of variation of the constant*. Consider first corresponding linear homogeneous equation\begin{equation}

y'= -\frac{y}{x}.

\label{2}

\end{equation}

It has a solution

\begin{equation}

y= Cx^{-1}

\label{3}

\end{equation}

with constant $C$ (do it by yourself!). Now consider (\ref{3}) with $C$ which is not a constant (variation!) and plug it into (\ref{1}).

\begin{gather*}

(Cx^{-1})' =-x^{-1}(Cx^{-1})+(Cx^{-1})^2\implies C'x^{-1}-Cx^{-2}=-Cx^{-2} + C^2x^{-2}\implies C' =C^2x^{-1}\\

\implies \frac{dC}{C^2}=\frac{dx}{x}\implies -\frac{1}{C}=\ln(x)+c

\end{gather*}

where $c=\mathsf{const}$. Then $C=-\dfrac{1}{\ln (x)+c}$ and plugging into (\ref{3}) we get

\begin{equation*}

\boxed{ y=-\frac{1}{x(\ln (x)+c)}.}

\end{equation*}

**B.**Consider equation\begin{equation}

y'- \tan(x) y =\cos(x).

\label{4}

\end{equation}

We solve it by

*the method of integrating factor*. Multiplying (\ref{4}) by unknown yet factor $\mu=\mu(x)$ we get\begin{equation}

\mu y' - \mu \tan (x) y = \mu \cos(x).

\label{5}

\end{equation}

*We want*the left hand expression to be $\mu y'+\mu 'y$, which means\begin{equation*}

\mu'=-\mu \tan(x) \implies \frac{d\mu}{\mu}= \tan(x)\,dx \implies \ln (\mu) =-\int \tan(x)\,dx = -\int\frac{\sin(x)}{\cos(x)}\,dx = \ln (\cos(x)).

\end{equation*}

You

**must**know this integral. We do not need any constant since we need just one integrating factor.So $\mu =\cos(x)$. (\ref{5}) is now

\begin{gather*}

\cos(x)y'-\sin (x)y =\cos^2(x)\implies \bigl(\cos(x)y\bigr)'= \cos^2(x)\implies

\cos(x)y = \int \cos^2(x)\,dx = \int \frac{1+\cos(2x)}{2}\,dx =\\

\frac{x}{2}+\frac{\sin (2x)}{4}+C=

\frac{x}{2}+\frac{\sin (x)\cos(x)}{2}+C.

\end{gather*}

You

**must**know simple trigonometric formulae. Then\begin{equation*}

\boxed{y= \frac{1}{2}(x+2C)\sec(x)+ \frac{1}{2}\sin (x).}

\end{equation*}