forked from Sasserisop/MATH201
546 lines
11 KiB
HTML
546 lines
11 KiB
HTML
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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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imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!
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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 …">
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<meta property="og:description" content="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!
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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)
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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!" />
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<meta name="twitter:description" content="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!
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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)
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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!"/>
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<h1 id="two-new-equations">two new equations</h1>
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<h2 id="linear-coefficients-equations">Linear coefficients equations</h2>
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<p>$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$
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imagine $c_{1},c_{2}=0$ It becomes a homogenous equation!</p>
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<p>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
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$(a_{1}u+b_{1}vK+\underbrace{\cancel{ c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l } }_{ 0 })dv=0$
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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 $\det(a_{1},b_{1},a_{2},b_{2})\ne 0$ turn into homogenous
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if $\det(\dots)=0 \Rightarrow$ equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)</p>
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<h3 id="example">Example</h3>
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<p><a class="hashtag" onclick="focusTag(this)">ex</a> <a class="hashtag" onclick="focusTag(this)">de_LC_type1</a>
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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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$(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$
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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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$det(-3,1,1,1)=-4$ //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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$(-3u+v)du+(u+v)dv=0$ //Beutiful1! it’s homogenous now
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$\frac{ dv }{ du }=\frac{{3u-v}}{u+v}=\frac{{3-\frac{v}{u}}}{1+\frac{v}{u}}$
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$\frac{v}{u}=w \quad v=uw \quad \frac{ dv }{ du }=w+u\frac{ dw }{ du }$
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$w+u\frac{ dw }{ du }=\frac{{3-w}}{1+w}$ This is the equation we have to solve
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$u\frac{ dw }{ du }=\frac{{3-2w-w^2}}{1+w}$
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$-\frac{{w+1}}{w^2+2w-3}dw=\frac{du}{u}$
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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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$\frac{1}{2}\int \frac{dz}{z}=\ln\mid u\mid^{-1}$
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$\ln\mid z\mid^{1/2}-\ln\mid u\mid^{-1}=C$
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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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$\mid z\mid u^2=e^{2C}$
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$zu^2=A$
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$\left( \left( \frac{v}{u} \right)^2+\frac{2v}{u}-3 \right)u^2=A$
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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: $(y+3)^2+2(y+3)(x-1)-3(x-1)^2=A$</p>
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<hr>
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<h2 id="exact-equations">Exact equations</h2>
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<p>two variable equations
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$dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ suppose it equals to zero (as shown in the equation) you get a horizontal plane (a constant)
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so $F(x,y)=C$
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the solution to these exact equations is given by $F()$ but how do we get F from the derivatives?
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Equation of the form: $$M(x,y)dx=N(x,y)dy=0$$
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is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$
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then differentiating we get:
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$\frac{ \partial M }{ \partial y }=\frac{ \partial^{2} F }{ \partial y\partial x }$
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$\frac{ \partial N }{ \partial x }=\frac{ \partial^{2} F }{ \partial x\partial y }$ 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)
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Exact equation$\Rightarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ if it’s continuous (?)
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also: Exact equation$\Leftarrow \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$
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Test for exactness:
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exact $\iff \frac{ \partial M }{ \partial y }=\frac{ \partial N }{ \partial x }$ (this can be proved, but it wasnt proved in class)
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<a class="hashtag" onclick="focusTag(this)">end</a> of lecture 4</p>
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<h3>Referenced in</h3>
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<ul>
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<li>No backlinks found</li>
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</ul>
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</main>
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