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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.
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*}
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*}