MATH 308 Lecture 8

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Lecture Notes

Autonomous Equations

$y'(x)=f(y)=y^{2}\left(y^{2}-4\right)$

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 $y''$ ${\begin{aligned}y'(x)&=f(y(x))\\y''(x)&=f'(y)\,y'(x)=f'(x)\,f(y)\end{aligned}}$

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 $y\,\mathrm {e} ^{x\,y}+x\,y'\,\mathrm {e} ^{x\,y}+2x=0$?

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

${\begin{aligned}{\frac {\mathrm {d} F}{\mathrm {d} x}}={\frac {\partial F}{\partial x}}\,{\frac {\mathrm {d} x}{\mathrm {d} x}}+{\frac {\partial F}{\partial y}}\,{\frac {\mathrm {d} y}{\mathrm {d} x}}&=0\\{\frac {\partial F}{\partial x}}+{\frac {\partial F}{\partial y}}\,y'&=0\end{aligned}}$

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

${\begin{aligned}{\frac {\partial F}{\partial y}}&=x\,\mathrm {e} ^{x\,y}\\F(x,y)&=\mathrm {e} ^{x\,y}\\{\frac {\partial F}{\partial x}}&=2x+y\,\mathrm {e} ^{x\,y}\\&=y\,\mathrm {e} ^{x\,y}+k'(x)\\k'(x)=2x\\k(x)=x^{2}+C\end{aligned}}$

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

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

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

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

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

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

Exercise 4

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

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

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:

${\begin{aligned}{\frac {1}{x\,y^{3}}}\left((x^{2}\,y^{2})\,\mathrm {d} x+x(1+y^{2})\,\mathrm {d} y\right)&=0\\x\,\mathrm {d} x+{\frac {1+y^{2}}{y^{3}}}\,\mathrm {d} y&=0\end{aligned}}$

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

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