Question

check my work

Answer 1

solve:
\[\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}\]
Let
\[x=X+h \space \space ; \space \space y=Y+k\]
\[\therefore \frac{dY}{dX}=\frac{2X+9Y+2h+9k-20}{6X+2Y+6h+2k-10}\]
\[3(2h+9k-20)=6h+27k-60\]
\[(6h+27k)-(6h+2k)=60-10\]\[25k=50 \implies k=2\]\[h=\frac{20-9k}{2}=\frac{20-18}{2}=\frac{2}{2}=1\]
\[\therefore X=x-1\]\[\therefore Y=y-2\]
\[\frac{dY}{dX}=\frac{2X+9Y}{6X+2Y}=\frac{2+9\frac{Y}{X}}{6+\frac{Y}{X}}\]
substitute \[Y=vX\]\[v+X\frac{dv}{dX}=\frac{2+9v}{6+2v}\]
\[X\frac{dv}{dX}=\frac{2+9v-6v-2v^2}{6+2v}=\frac{-2v^2+3v+2}{6+2v}\]
\[\frac{6+2v}{-2v^2+3v+2}.dv=\frac{dX}{X}\]
Let
\[6+2v=A \frac{d}{dv}(-2v^2+3v+2)+B\]\[6+2v=A(-4v+3)+B\]\[6+2v=(3A+B)-4Av\]
Comparing coefficients
\[-4A=2 \space \space \therefore \space \space A=-\frac{1}{2}\]
and
\[-\frac{3}{2}+B=6 \space \space \therefore \space \space B=6+\frac{3}{2}=\frac{15}{2}\]
\[\therefore -\frac{1}{2}\frac{(-4v+3)}{(-2v^2+3v+2)}.dv+\frac{15}{2}\frac{dv}{-2v^2+3v+2}=\frac{dX}{X}\]
and we have
\[-2v^2+3v+2=-2(v^2-\frac{3}{2}v-1)\]\[-2v^2+3v+2=-2(v^2-2.\frac{3}{4}v+\frac{9}{16}-\frac{25}{16})\]
\[-2v^2+3v+2=-2((v-\frac{3}{4})^2-(\frac{5}{4})^2)\]
\[\therefore -\frac{1}{2}\frac{(-4v+3)}{(-2v^2+3v+2)}.dv-\frac{15}{4}\frac{dv}{(v-\frac{3}{4})^2-(\frac{5}{4})^2}=\frac{dX}{X}\]
\[-\frac{1}{2}\int\limits \frac{f'(v)}{f(v)}dv-\frac{15}{4}\int\limits \frac{dt}{t^2-(\frac{5}{4})^2}=\int\limits \frac{dX}{X}\]
\[\int\limits \frac{dx}{x^2-a^2}=\frac{1}{2a}\log|\frac{x-a}{x+a}|+C\]
\[-\frac{1}{2}\log|-2v^2+3v+2|-\frac{15}{4}.\frac{5}{2 \times 5}\log|\frac{v-\frac{4}{8}}{v+\frac{2}{4}}|=\log|CX|\]
\[-\frac{1}{2}\log|-2((v-\frac{3}{4})^2-(\frac{5}{4})^2)|-\frac{3}{2}\log|\frac{v-2}{v+\frac{1}{2}}|=\log|CX|\]\[\log|-2((v-\frac{3}{4}-\frac{5}{4})(v-\frac{3}{4}+\frac{5}{4}))|+3\log|2\frac{v-2}{2v+1}|=\log|\frac{1}{C^2X^2}|\]\[4\log2+\log|(v-\frac{8}{4})(v+\frac{1}{2})|+3\log|\frac{v-2}{2v+1}|=\log \frac{1}{C^2X^2}\]\[4\log2+\log|\frac{(v-2)(2v+1)}{2}|+\log|(\frac{v-2}{2v+1})^3|=\log \frac{1}{C^2X^2}\]\[3\log2+\log|(v-2)(2v+1)|+\log|(\frac{v-2}{2v+1})^3|=\log \frac{1}{C^2X^2}\]
\[\log|\frac{(v-2)^4}{(2v+1)^2}|=\log \frac{1}{C^2X^2}-\log2^3\]
\[\log \frac{(v-2)^2}{(2v+1)^2}=\log \frac{1}{2^3C^2X^2}\]\[\frac{(v-2)^4}{(2v+1)^2}=\frac{1}{2^3C^2X^2}\]\[\frac{(v-2)^2}{2v+1}=\frac{1}{2\sqrt{2}CX}\]
\[\frac{\frac{(Y-2X)^2}{X^2}}{\frac{2Y+X}{X}}=\frac{1}{2\sqrt{2}CX}\]
\[\frac{(Y-2X)^2}{2Y+X}=C' \space \space ; \space \space C'=\frac{1}{2\sqrt{2}C}\]
\[\frac{(y-2-2(x-1))^2}{2(y-2)+x-1}=C'\]\[(y-2-2x+2)^2=(2y-4+x-1)C'\]\[(2x-y)^2=(x+2y-5)C'\]

Answer 2

100%

Answer 3

no jk

Answer 4

@ganeshie8 @IrishBoy123