forked from Sasserisop/MATH201
68 lines
2.8 KiB
Markdown
68 lines
2.8 KiB
Markdown
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#start of lecture 4
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## Linear coefficients equations
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$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$
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> I'm calling this #de_LC_type1
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imagine $c_{1},c_{2}=0$ It becomes a homogenous equation! #de_h_type2
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so can we make them 0?
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let $x=u+k$
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$y=v+l$
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where $k,l$ are constants hand picked such that the following terms equal 0:
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$(a_{1}u+b_{1}v+\underbrace{\cancel{ c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l } }_{ 0 })dv=0$
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In order for these two terms to equal zero, we have to solve this linear system of equations:
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$a_{1}k+b_{1}l=-c_1$
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$a_{2}k+b_{2}l=-c_{2}$
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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.
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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)
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### Example
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#ex #de_LC_type1
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$$(-3x+y+6)dx+(x+y+2)dy=0$$
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let $x=u+k$
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$y=v+l$
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differentiating we get: $dx=du ,\quad dy=dv$
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$(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$
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we want $6-3k+l$ and $2+k+l$ to equal 0
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so:
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$-3k+l=-6$
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$k+l=-2$
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$det\begin{pmatrix}-3 & 1 \\1 & 1\end{pmatrix}=-4$ //you call it a fish? He can call it a dinosaur if he wanted to :D
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solving gives us:
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$k=1,l=-3$
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so $x=u+1 \quad y=v-3$
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$(-3u+v)du+(u+v)dv=0$ //Beautiful! It's homogenous now
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$\frac{ dv }{ du }=\frac{{3u-v}}{u+v}$
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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.
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$\frac{ dv }{ du }=\frac{{3-\frac{v}{u}}}{1+\frac{v}{u}}$
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$\frac{v}{u}=w \quad v=uw \quad \frac{ dv }{ du }=w+u\frac{ dw }{ du }$
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$w+u\frac{ dw }{ du }=\frac{{3-w}}{1+w}$ 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:
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$u\frac{ dw }{ du }=\frac{{3-2w-w^2}}{1+w}$
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$-\frac{{w+1}}{w^2+2w-3}dw=\frac{du}{u}$ <- Like that :)
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$\int-\frac{{w+1}}{w^2+2w-3}dw=\int\frac{du}{u}$
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let $z=w^2+2w-3$
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$dz=2(w+1)dw$
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$\frac{-1}{2}\int \frac{dz}{z}=\ln\mid u\mid$
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$\frac{-1}{2}\ln|z|+C=\ln\mid u\mid$
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$\ln\mid z\mid^{1/2}+\ln\mid u\mid=C$
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$\ln(\mid z\mid^{1/2}\mid u\mid)=C$
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$\mid z\mid^{1/2}u=e^C$
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> How did he get rid of the abs()? I'm not sure. But he fixes the problem right after:
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$\mid z\mid u^2=e^{2C}$
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> 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.
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$zu^2=A$
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> This step I can understand.
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$\left( \left( \frac{v}{u} \right)^2+2\frac{v}{u}-3 \right)u^2=A$
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remember $u=x-1 \quad v=y+3$
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$$\left( \left( \frac{{y+3}}{x-1} \right)^2+\frac{2(y+3)}{x-1}-3 \right)(x-1)^2=A$$
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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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---
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