MATH 308 Lecture 9

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Lecture Notes

Exact Equations

$(\frac{\sin y}{y}-2{\mathrm{e}}^{-x}\sin x)\mathrm{d}x+\left(\frac{\cos y+2{\mathrm{e}}^{-x}\cos x}{y}\right)\mathrm{d}y=0$

Recall that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ in order for the differential equation to be considered exact

$\begin{array}{rl}\frac{\partial M}{\partial y}& =\frac{\cos y}{y}-\frac{\sin y}{y^2}\\ \frac{\partial N}{\partial x}& =\frac{2}{y}(-{\mathrm{e}}^{-x}\cos x-\sin x{\mathrm{e}}^{-x})\end{array}$

Equation is not exact so we can stop here.

We multiply by the integrating factor $\mu (x,y)=y{\mathrm{e}}^x$ to obtain

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=\cos y{\mathrm{e}}^x-2\sin x$

Now we differentiate $F(x,y)=(\sin y{\mathrm{e}}^x-2y\sin x)\mathrm{d}x+(\cos y{\mathrm{e}}^x+2\cos x)\mathrm{d}x=0$ with respect to $y$

$\begin{array}{rl}\frac{\partial F}{\partial y}& =\cos y{\mathrm{e}}^x+2\cos x+k^{\prime }(y)\\ \dots \end{array}$

Implicit solution: $\sin y{\mathrm{e}}^x+2y\cos x+C=0$

$y^{\prime }(t)=f(y)={\mathrm{e}}^y-1$

Equilibrium solution:

$\begin{array}{rl}f(y)& =0\\ {\mathrm{e}}^y-1& =0\\ {\mathrm{e}}^y& =1\\ y& =\ln 1\\ y& =0\end{array}$

Is it stable/unstable/semi-stable?

• for $y<0$, the function is decreasing.
• for $y>0$, the function is increasing.
• unstable.

Phase line:

 -----------<-----------|----------->-----------
          y < 0         0         y > 0
         y' < 0 (dec)            y' > 0 (inc)
        y'' < 0 (ccdn)          y'' > 0 (ccup) 

Concavity?

$\begin{array}{rl}{y}^{\prime }& =f(y)\\ y^{″}& =f^{\prime }(y)f(y)\\ f^{\prime }(y)& ={\mathrm{e}}^y>0\end{array}$

Linear vs Nonlinear

Linear	non-linear
$y^{\prime }+P(x)y=g(x)y(x_0)=y_0$	$y^{\prime }=f(x,y)y(x_0)=y_0$
$P(x)$ and $g(x)$ are continuous on an open interval $I$ $x_0\in I$, $y_0\in \mathbb{ℝ}$	$f(x,y)$ and $\frac{\partial f}{\partial y}$ are continuous on $I\times J$ (product of 2 intervals) $x_0\in I$, $y_0\in J$
solution to init. val. problem exists solution is unique solution exists on $I$	solution to init. val. problem exists solution is unique solution exists on interval included in $I$

Exercise 13

$y^{\prime }(t)=\frac{t-y}{2t+5y}y(t_0)=y_0$

nonlinear, so $f(t,y)=\frac{t-y}{2t+5y}$ exists on $2t+5y\ne 0$

Solution does not exist on line $y=-\frac{2}{5}t$

Exercise 12

$(t-3)y^{\prime }+y\ln t=2ty(1)=2$

Linear diff eq. in "standard" form:

$y^{\prime }+\frac{\ln t}{t-3}y=2t$

Function $P(t)=\frac{\ln t}{t-3}$ exists on $t\ne 3$ and $t>0$: $t\in (0,3)\cup (3,\mathrm{\infty })$

There exists a unique solution to the initial value problem $y(1)=2$ and the domain of the solution is on (0, 3).