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Question Number 169558 by greougoury555 last updated on 03/May/22

$$\frac{dy}{dx}=\frac{2x-y+1}{x-4y+3}$$

Answered by ali009 last updated on 03/May/22

$$a1=2a2=1a1/a2=2$$$$b1=-1b2=-4b1/b2=\frac{1}{4}$$$$so$$$$2x-y+1=0$$$$-2x+8y-6=0$$$$-------$$$$7y-5=0$$$$y=\frac{5}{7}$$$$\frac{5}{7}=2x+1$$$$x=-\frac{1}{7}$$$$letx=X-\frac{1}{7}dx=dX$$$$y=Y+\frac{5}{7}dy=dY$$$$\frac{dY}{dX}=\frac{2(X-\frac{1}{7})-(Y+\frac{5}{7})+1}{(X-\frac{1}{7})-4(Y+\frac{5}{7})+3}$$$$\frac{dY}{dX}=\frac{2X-Y}{X-4Y}$$$$\frac{dY}{dX}=\frac{2-\frac{Y}{X}}{1-4\frac{Y}{x}}$$$$let\frac{Y}{X}=v$$$$$$$$\frac{dY}{dX}=v+X\frac{dv}{dX}$$$$v+X\frac{dv}{dX}=\frac{2-v}{1-4v}$$$$X\frac{dv}{dX}=\frac{2-v}{1-4v}-v$$$$X\frac{dv}{dX}=\frac{2-v-(1-4v)\left(v\right)}{1-4v}$$$$X\frac{dv}{dX}=\frac{2-v-v+4v^2}{1-4v}$$$$\frac{1-4v}{4v^2-2v+2}dv=\frac{dx}{X}$$$$-\frac{1}{2}\int \frac{4v-1}{2v^2-v+2}dv=\int \frac{dx}{X}$$$$-\frac{1}{2}ln(2v^2-v+2)=ln\left(X\right)+c$$$$ln(2v^2-v+2)=-ln\left(X\right)+C_1$$$$2v^2-v+2=\frac{C_2}{X}$$$$2{\left(\frac{Y}{X}\right)}^2-\frac{Y}{x}+2=\frac{C_2}{X}$$$$(x+\frac{1}{7})(2{\left(\frac{y-\frac{5}{7}}{x+\frac{1}{7}}\right)}^2-\frac{y-\frac{5}{7}}{x+\frac{1}{7}}+2)=C_2$$$$$$

Commented by Tawa11 last updated on 03/May/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 03/May/22

$$letx=u+a,y=v+b$$$$2x-y+1=2u-v+(2a-b+1)$$$$x-4y+3=u-3b+(a-4b+3)$$$$set$$$$2a-b+1=0$$$$a-4b+3=0$$$$i.e.a=-\frac{1}{7},b=\frac{5}{7}$$$$\Rightarrow \frac{dy}{dx}=\frac{2u-v}{u-4v}$$$$\Rightarrow \frac{dv}{du}=\frac{2u-v}{u-4v}$$$$letv=ut$$$$\frac{dv}{du}=t+u\frac{dt}{du}$$$$\Rightarrow t+u\frac{dt}{du}=\frac{2-t}{1-4t}$$$$\Rightarrow u\frac{dt}{du}=\frac{2(1-t+2t^2)}{1-4t}$$$$\Rightarrow \frac{(4t-1)dt}{2t^2-t+1}=-\frac{2du}{u}$$$$\Rightarrow \int \frac{(4t-1)dt}{2t^2-t+1}=-\int \frac{2du}{u}$$$$\Rightarrow \int \frac{d(2t^2-t+1)}{2t^2-t+1}=-\int \frac{2du}{u}$$$$\Rightarrow \ln (2t^2-t+1)=-2\ln u+C$$$$(2t^2-t+1)u^2=C$$$$(\frac{2v^2}{u^2}-\frac{v}{u}+1)u^2=C$$$$[2{\left(\frac{y+\frac{5}{7}}{x-\frac{1}{7}}\right)}^2-\left(\frac{y+\frac{5}{7}}{x-\frac{1}{7}}\right)+1]{(x-\frac{1}{7})}^2=C$$$$\Rightarrow [2{\left(\frac{7y+5}{7x-1}\right)}^2-\frac{7y+5}{7x-1}+1]{(7x-1)}^2=C$$

Commented by Tawa11 last updated on 03/May/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by princeDera last updated on 15/Aug/22

$$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{2\mathit{x}-\mathit{y}+1}{\mathit{x}-4\mathit{y}+3}$$$$$$$$(\mathit{x}-4\mathit{y}+3)\mathit{d}\mathit{y}+(\mathit{y}-2\mathit{x}-1)\mathit{d}\mathit{x}=0$$$$$$$$\mathit{m}=\mathit{x}-4\mathit{y}+3\mathit{a}\mathit{n}\mathit{d}\mathit{n}=\mathit{y}-2\mathit{x}-1$$$$\frac{\partial }{\partial \mathit{x}}\mathit{m}=1\mathit{a}\mathit{n}\mathit{d}\frac{\partial }{\partial \mathit{y}}\mathit{n}=1$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{D}.\mathit{E}\mathit{i}\mathit{s}\mathit{e}\mathit{x}\mathit{a}\mathit{c}\mathit{t}$$$$$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{i}\mathit{s}\mathit{o}\mathit{f}\mathit{t}\mathit{h}\mathit{e}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{f}(\mathit{x},\mathit{y})=\mathit{c}$$$$\frac{\partial }{\partial \mathit{x}}\mathit{f}(\mathit{x},\mathit{y})=\mathit{y}-2\mathit{x}-1$$$$\mathit{f}(\mathit{x},\mathit{y})=\int (\mathit{y}-2\mathit{x}-1)\mathit{d}\mathit{x}=\mathit{x}\mathit{y}-{\mathit{x}}^2-\mathit{x}+\mathit{h}\left(\mathit{y}\right)$$$$$$$$\frac{\partial }{\partial \mathit{y}}\mathit{f}(\mathit{x},\mathit{y})=\mathit{x}+{\mathit{h}}^{\prime }\left(\mathit{y}\right)=\mathit{x}-4\mathit{y}+3$$$${\mathit{h}}^{\prime }\left(\mathit{y}\right)=3-4\mathit{y}$$$$\mathit{h}\left(\mathit{y}\right)=3\mathit{y}-2{\mathit{y}}^2$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{t}\mathit{o}\mathit{t}\mathit{h}\mathit{e}\mathit{D}.\mathit{E}=\mathit{f}(\mathit{x},\mathit{y})=\mathit{c}$$$$\mathit{x}\mathit{y}-{\mathit{x}}^2-\mathit{x}+3\mathit{y}-2{\mathit{y}}^2=\mathit{c}$$$$$$$$$$$$$$$$$$

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