MATH 308 Lecture 8

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Lecture Notes

Autonomous Equations

$y^{\prime }(x)=f(y)=y^2(y^2-4)$

Solving $f(y)=0$ gives $y\in \{-2,0,2\}$, which are equilibrium solutions.

• $y=2$ is an unstable equilibrium solution since any deviation around it will "push away"
• $y=0$ is a semistable equilibrium solution since one side of it (towards $y=2$) pushes towards it and the other side pushes away
• $y=-2$ is a stable equilibrium solution since either side converges to it.

Concavity

Second derivative test: What is $y^{″}$ $\begin{array}{rl}{y}^{\prime }(x)& =f(y(x))\\ y^{″}(x)& =f^{\prime }(y)y^{\prime }(x)=f^{\prime }(x)f(y)\end{array}$

Recall that when $y^{″}>0$, the function $y$ is concave up and when $y^{″}<0$, the function is concave down.

Exact Equations and Integrating Factors

Can we find an implicit expression for the solution to $y{\mathrm{e}}^{xy}+x{y}^{\prime }{\mathrm{e}}^{xy}+2x=0$?

Expression with $x$ and $y$: $F(x,y(x))=0$, but what is $F(x,y(x))$?

$\begin{array}{rl}\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\partial F}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{\partial F}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}& =0\\ \frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}{y}^{\prime }& =0\end{array}$

It would be great if $\frac{\partial F}{\partial x}=y{\mathrm{e}}^{xy}+2x$ and $\frac{\partial F}{\partial y}=x{\mathrm{e}}^{xy}$.

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

Theorem

Given an equation

$M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0$

If $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, then there exists a function $F$ such that

$\frac{\partial F}{\partial x}=M$ and $\frac{\partial F}{\partial y}=N$

Exercise 2

Is the equation $(2xy+3)\mathrm{d}x+(x^2-1)\mathrm{d}y=0$ exact?

Let $M(x,y)=2xy+3$ and $N(x,y)=x^2-1$.

$\begin{array}{rl}\frac{\partial M}{\partial y}& =2x\\ \frac{\partial N}{\partial x}& =2x\end{array}$

According to the #Theorem, this equation is exact, and $F(x,y)=0$ is an implicit solution.

$\begin{array}{rl}\frac{\partial F}{\partial x}& =2xy+3\\ F(x,y)& =x^{2}y+3x+k(y)\\ \frac{\partial F}{\partial y}& =x^2+k^{\prime }(y)\\ & =x^2-1\\ k^{\prime }(y)& =-1\\ k(y)& =-y+C\end{array}$

Therefore, $F(x,y)=x^{2}y+3x-y+C$

Exercise 4

Is the equation $x^2{y}^3+x(1+y^2)y^{\prime }$ exact?

$\begin{array}{rl}\frac{\partial M}{\partial y}& =3x^2{y}^3\\ \frac{\partial N}{\partial x}& =1+y^2\end{array}$

They are not equal, so it is not exact.

However, if we multiply by the integrating factor $\frac{1}{x{y}^2}$, it is solvable:

$\begin{array}{rl}\frac{1}{x{y}^3}((x^2{y}^2)\mathrm{d}x+x(1+y^2)\mathrm{d}y)& =0\\ x\mathrm{d}x+\frac{1+y^2}{y^3}\mathrm{d}y& =0\end{array}$

Now the equation is exact since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=0$

$\begin{array}{rl}\frac{\partial F}{\partial x}& =x\\ F(x,y)& =\frac{x^2}{2}+k(y)\\ \frac{\partial F}{\partial y}& =0+k^{\prime }(y)\\ & =\frac{1}{y^3}+\frac{1}{y}\\ k(y)& =\frac{1}{-2y^2}+\ln |y|+C\end{array}$