two new equations

Linear coefficients equations

$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$ imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!

so can we make them 0? let $x=u+k$ $y=v+l$ where $k,l$ are constants $(a_{1}u+b_{1}vK+\underbrace{\cancel{ c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l } }_{ 0 })dv=0$ $a_{1}k+b_{1}l=-c_1$ $a_{2}k+b_{2}l=-c_{2}$ if $\det(a_{1},b_{1},a_{2},b_{2})\ne 0$ turn into homogenous if $\det(\dots)=0 \Rightarrow$ equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)

Example

ex de_LC_type1 $$(-3x+y+6)dx+(x+y+2)dy=0$$ let $x=u+k$ $y=v+l$ $(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$ we want $6-3k+l$ and $2+k+l$ to equal 0 so: $-3k+l=-6$ $k+l=-2$ $det(-3,1,1,1)=-4$ //he can call it a dinosaur if he wanted to :D solving gives us: $k=1,l=-3$ so $x=u+1 \quad y=v-3$ $(-3u+v)du+(u+v)dv=0$ //Beutiful1! it’s homogenous now $\frac{ dv }{ du }=\frac{{3u-v}}{u+v}=\frac{{3-\frac{v}{u}}}{1+\frac{v}{u}}$ $\frac{v}{u}=w \quad v=uw \quad \frac{ dv }{ du }=w+u\frac{ dw }{ du }$ $w+u\frac{ dw }{ du }=\frac{{3-w}}{1+w}$ This is the equation we have to solve $u\frac{ dw }{ du }=\frac{{3-2w-w^2}}{1+w}$ $-\frac{{w+1}}{w^2+2w-3}dw=\frac{du}{u}$ $\int-\frac{{w+1}}{w^2+2w-3}dw=\int\frac{du}{u}$ let $z=w^2+2w-3$ $dz=2(w+1)dw$ $\frac{1}{2}\int \frac{dz}{z}=\ln\mid u\mid^{-1}$ $\ln\mid z\mid^{1/2}-\ln\mid u\mid^{-1}=C$ $\ln(\mid z\mid^{1/2}\mid u\mid)=C$ $\mid z\mid^{1/2}u=e^C$ $\mid z\mid u^2=e^{2C}$ $zu^2=A$ $\left( \left( \frac{v}{u} \right)^2+\frac{2v}{u}-3 \right)u^2=A$ remember $u=x-1 \quad v=y+3$ $$\left( \left( \frac{{y+3}}{x-1} \right)^2+\frac{2(y+3)}{x-1}-3 \right)(x-1)^2=A$$ you can “simplify” it to: $(y+3)^2+2(y+3)(x-1)-3(x-1)^2=A$

Exact equations

two variable equations $dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ suppose it equals to zero (as shown in the equation) you get a horizontal plane (a constant) so $F(x,y)=C$ the solution to these exact equations is given by $F()$ but how do we get F from the derivatives? Equation of the form: $$M(x,y)dx=N(x,y)dy=0$$ is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$ then differentiating we get: $\frac{ \partial M }{ \partial y }=\frac{ \partial^{2} F }{ \partial y\partial x }$ $\frac{ \partial N }{ \partial x }=\frac{ \partial^{2} F }{ \partial x\partial y }$ Order of going in x then y vs y then x doesn’t matter as it lands you on the same point (idk how this is related yet) Exact equation$\Rightarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ if it’s continuous (?) also: Exact equation$\Leftarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ Test for exactness: exact $\iff \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ (this can be proved, but it wasnt proved in class) end of lecture 4

Referenced in