Date modified: Mon 2025-06-30 . 11:30 AM
Date created: Sat 2025-06-28 . 11:31 AM
# Week 2 Monday Let us see some application of the ideas we've learned. As well some additional techniques of dealing with first-order ODEs. ## Orthogonal family of curves. - Given a $1$-parameter family of curves, $\cal F$, can we find another $1$-parameter family of curves $\cal G$, such that each curve in $\cal F$ is perpendicular (orthogonal) to each curve in $\cal G$, whenever they intersect. - Such a family $\cal G$ is said to be an **orthogonal family** of curves to $\cal F$. - In physics, sometimes we find these orthogonal family of curves, like electric field and electric potential. - For example, say ${\cal F}= \{4x+3y=C : C \text{ free}\}$, all lines of slope $-4 / 3$. Then ${\cal G}=\{ 3x - 4y = C : C \text{ is free} \}$ consists of all lines of slope $3/4$. - When this happens, at the point of intersection, the curve from $\cal F$ must have a slope (derivative) that is negative reciprocal of the slope of the curve in $\cal G$. - Recall we can describe a 1-parameter family of curves $\cal F$ with a potential function $\phi(x,y)$, in particular ${\cal F }= \{\phi(x,y) = C : C \text{ free} \}$. So the curves in $\cal F$ must satisfy $\mathrm{d}\phi =0$, or $\phi_{x}\mathrm{d}x + \phi_{y} \mathrm{d}y=0$. So the family $\cal F$ has curves that satisfy the slope field $\frac{\mathrm{d}y}{\mathrm{d}x}=- \frac{\phi_{x}}{\phi_{y}}$. - As $\cal G$ is orthogonal to $\cal F$, the curves in $\cal G$ satisfy the slope field $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\phi_{y}}{\phi_{x}}$. So all we need is to solve this differential equation! In other words, $\cal G$ has curve satisfying $\phi_{y}\mathrm{d}x -\phi_{x} \mathrm{d}y=0$. - Basic idea: Find out the slope field of $\cal F$, then take the negative reciprocal to get the slope field for $\cal G$. - Example. Let $\cal F$ be the set of all parabolas of the form $y = C x^{2}$. Find a family of curves $\cal G$ orthogonal to $\cal F$. - First let us find the slope field of $\cal F$. It's easiest by describing the curves in $\cal F$ in the form of potential $\phi = C$. So here we have $\phi = y / x^{2} = C$. - So the DE satisfied by $\cal F$ is $\mathrm{d}\phi = 0$, namely $- 2 y / x ^3 \mathrm{d}x + 1 / x^{2} \mathrm{d}y = 0$. - And the DE satisfied by $\cal G$ is then $1 / x ^ 2 \mathrm{d}x + 2y / x^{3} \mathrm{d}y=0$. Though this is not exact, it is in fact separable: $\frac{\mathrm{d}y}{\mathrm{d}x}= - \frac{x}{2y}$. So $2y \mathrm{d}y = -x\mathrm{d}x$, and integrating both sides give $y^{2}=-\frac{x^{2}}{2} + C$. In other words, $\frac{x^{2}}{2}+y^{2}=C$, which are ellipses. - Plot these on Desmos. - Example. How about this family $\cal F$ of parabola: $y = x^{2} + C$? Find a family of curves $\cal G$ orthogonal to $\cal F$. - Note the potential for $\cal F$ is $\phi = y-x^{2}= C$. So $\cal F$ has curves satisfying $-2x \mathrm{d}x + \mathrm{d}y = 0$. - So curves in $\cal G$ satisfy $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-1}{2x}$, which integrates to $y = -\frac{1}{2}\ln |x| + C$. ## Substitution methods. The oldest trick in the book. Substitution sometimes make problem easier. The idea is that upon substitution, we can something that is more familiar that we can handle. Let us explore a few cases. ### Secret lower order. - Sometimes a higher order problem is secretly a lower order problem. - Try $w = y^{(n)}$ for some suitable $n$. - Example. Solve for $y=y(x)$ such that $y''' + 2 y'' = x$. - Answer. Set $w = y''$. Then this DE becomes $w'+2w =x$. This is linear! We have integrating factor $\mu = e^{\int2dx}= e^{2x}$, so we can write down $w=e^{-2x}\left[\int e^{2x} x dx+C \right] =$ $e^{-2x} \left[\int x d \left( \frac{1}{2}e^{2x} \right) +C \right] =$ $e^{-2x} \left[ \frac{1}{2}xe^{2x} - \frac{1}{4} e^{2x} +C\right]=\frac{1}{2}x- \frac{1}{4} + Ce^{-2x}$. - But we still have $w = y''$, now we use FTC to figure out $y$. - Integrating once gives $y' = \frac{1}{4}x^{2} - \frac{1}{4}x + Ce^{-2x}+ D$, abusing constants, but we do have a different constant of integration $D$ from $C$. - Integrating once more, gives $y = \frac{1}{12} x^{3} - \frac{1}{8} x^{2} +C e^{-2x} + Dx + E$. This is a 3-parameter family of solutions, which agrees with our original DE being third order. ### Linear substitution. - If we see terms like $Ax + By + C$ showing up repeatedly, or we have slope field $y'=f(Ax+By+C)$, then we can do a linear substitution. - Try $w = Ax + By + C$, where $A,B, C$ some constants, and $dw = Adx+Bdy$. - Example. Solve the DE $y' = \frac{1}{\sqrt{2x+3y+5}}$. Set $w = 2x + 3y +5$. Note then $dw = 2 dx + 3dy$. So we have $dy = \frac{1}{\sqrt{w}} dx$. And as $dy = \frac{1}{3}dw - \frac{2}{3}dx$, we have $dw -2dx = \frac{3}{\sqrt{w}} dx$ , so $dw = (2 + \frac{3}{\sqrt{w}})dx$, which is separable. This gives $\frac{\sqrt{w}}{2\sqrt{w}+3} dw = dx$. Integrate both sides. Which, incidentally, can use another substitution $u = 2\sqrt{w} + 3$. Finally, express everything back in $x,y$. ### Homogeneous of same degree. - A function $g(x_{1},x_{2},\ldots,x_{k})$ is said to be **homogeneous of degree $d$** if for all $x_{i}'s$, $g(\lambda x_{1}, \lambda x_{2}, \ldots, \lambda x_{k})=\lambda ^d g(x_1, x_{2}, \ldots, x_{k})$. That is, scaling all inputs by $\lambda$ is the same as scaling the overall function by $\lambda^{d}$, for any $\lambda$. - Example. $f(x,y,z) = 3x^{2}y - xyz + \frac{y^{5}}{xy}$ is homogeneous of degree 3 (over its domain). - Example. $f(x,y) = 5x^{2}-3xy$ is homogeneous of degree $2$, but $g(x,y) = 5x^{2}-3xy +2$ is not homogeneous of any degree. - Example. $f(x,y)=\sqrt{x+y}$ is homogeneous of degree $\frac{1}{2}$. - Suppose we have a differential equation of the form $M(x,y)dx + N(x,y)dy = 0$, where both $M(x,y)$ and $N(x,y)$ are **homogeneous of the same degree**, then use the substitution $y =wx$. The idea is $y$ and $x$ have the same degree contribution, so we can homogenize it and have $w = \frac{y}{x}$ (or its reciprocal). With $y = wx$, we have $dy = wdx+xdw$. The philosophy is $w$ is unit-less. Ignore this if it is confusing. - If successful, we will get a separable equation. - Example. $(x^{2}y+y^{3}) dx + x^{3}dy=0$. This is not exact, but it is homogeneous of degree $3$. - Set $y = wx$, and $dy = wdx + xdw$. Then we have $(x^{3}w+w^{3}x^{3})dx+x^{3}(wdx+xdw)=0$. Combining we get $(x^{3}w+x^{3}w^{3}+x^{3}w)dx + x^{4}dw = 0$, so $(2w+w^{3})dx+xdw = 0$. So we now have $$ \frac{dw}{w(2+w^{2})}=-\frac{1}{x}dx $$which is separable. To integrate this, you need partial fractions, and remember arctangent. - Partial fraction gives $\frac{1}{2w} - \frac{w}{2(w^{2}+2)} dw = -\frac{1}{x}dx$. You can integrate this. ### Bernoulli equations. - Sometimes it calls for curious substitutions. Here's a special kind that upon substitution reduces to a linear equation. - DE of the form $y'+P(x)y=Q(x)y^{n}$. Almost like linear except with power $y^{n}$. - Consider $w=y^{1-n}$. So $dw=(1-n)y^{-n}dy$. - Doing this turns the Bernoulli equation into something linear. - Example. $y'+2y=\frac{x}{y}$. This is Bernoulli, with $n=-1$. This means take $w = y^{1-(-1)}=y^{2}$, and $dw = 2ydy$. - Then we get $dy + 2y dx = \frac{x}{y}dx$, and $\frac{1}{2y}\frac{dw}{dx} + 2y= \frac{x}{y}$. Which is $w'+4y^{2}=2x$, or $w'+4w=2x$. This is linear! - Solve for $w$ using $P=4$ and $Q=2x$. This gives $\mu=e^{4x}$ and $w=e^{-4x}[\int 2xe^{4x}dx+C]$ = $\frac{1}{8}(4x-1) + Ce^{-4x}$. This gives implicit solution $y^{2}= \frac{1}{2}x-\frac{1}{8}+Ce^{-4x}$.