MATH201/content/Linear coefficient equations (lec 4).md

#start of lecture 4

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}

I'm calling this #de_LC_type1

imagine c_{1},c_{2}=0 It becomes a homogenous equation! #de_h_type2

so can we make them 0? let x=u+k y=v+l where k,l are constants hand picked such that the following terms equal 0: (a_{1}u+b_{1}v+\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 In order for these two terms to equal zero, we have to solve this linear system of equations: a_{1}k+b_{1}l=-c_1 a_{2}k+b_{2}l=-c_{2}

if \det\begin{pmatrix}a_{1} & b_{1} \\a_{2} & b_{2}\end{pmatrix}\ne0 the system is solvable and the DE turns into a homogenous equation. if \det\begin{pmatrix}a_{1} & b_{1} \\a_{2} & b_{2}\end{pmatrix}=0 \Rightarrow the system is unsolvable but we get an 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 differentiating we get: dx=du ,\quad dy=dv (-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\begin{pmatrix}-3 & 1 \\1 & 1\end{pmatrix}=-4 //you call it a fish? 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 //Beautiful! It's homogenous now \frac{ dv }{ du }=\frac{{3u-v}}{u+v} divide top and bottom by u so we turn the homogenous equation into the form #de_h_type1 and solve it using the tools we developed from lecture 2. \frac{ dv }{ du }=\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} If you remember from lecture 2, after these substitutions the equation should now be separable, we just move the w terms to one side and the u terms to the other: u\frac{ dw }{ du }=\frac{{3-2w-w^2}}{1+w} -\frac{{w+1}}{w^2+2w-3}dw=\frac{du}{u} <- Like that :) \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 \frac{-1}{2}\ln|z|+C=\ln\mid u\mid \ln\mid z\mid^{1/2}+\ln\mid u\mid=C \ln(\mid z\mid^{1/2}\mid u\mid)=C \mid z\mid^{1/2}u=e^C

How did he get rid of the abs()? I'm not sure. But he fixes the problem right after:

\mid z\mid u^2=e^{2C}

Funny enough, after that step above of squaring both sides is done, it's like he never even dropped the abs to begin with. All solutions are reobtained again.

zu^2=A

This step I can understand.

\left( \left( \frac{v}{u} \right)^2+2\frac{v}{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 But we are done.

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