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Question Number 12245 by tawa last updated on 16/Apr/17

$$\mathrm{Solve}\mathrm{the}\mathrm{differential}\mathrm{equation}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{x}+2\mathrm{y}-3}{8\mathrm{x}-4\mathrm{y}+5}$$

Answered by mrW1 last updated on 17/Apr/17

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{x}+2\mathrm{y}-3}{8\mathrm{x}-4\mathrm{y}+5}$$$$$$$$lety=u-p$$$$letx=v-q$$$$withp,q=constants$$$$dy=du$$$$dx=dv$$$$4(v-q)+2(u-p)-3=4v+2u+(-4q-2p-3)$$$$8(v-q)-4(u-p)+5=8v-4u+(-8q+4p+5)$$$$$$$$-4q-2p-3=0\left(i\right)$$$$-8q+4p+5=0\left(ii\right)$$$$\left(ii\right)-\left(i\right)\times 2:$$$$8p+11=0$$$$\Rightarrow p=-\frac{11}{8}$$$$\left(ii\right)+\left(i\right)\times 2:$$$$-16q-1=0$$$$\Rightarrow q=-\frac{1}{16}$$$$$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4\mathrm{x}+2\mathrm{y}-3}{8\mathrm{x}-4\mathrm{y}+5}\Rightarrow$$$$\frac{\mathrm{d}u}{\mathrm{d}v}=\frac{4v+2u}{8v-4u}$$$$letu=wv$$$$\frac{du}{dv}=w+v\frac{dw}{dv}$$$$$$$$\frac{\mathrm{d}u}{\mathrm{d}v}=\frac{4v+2u}{8v-4u}\Rightarrow$$$$w+v\frac{dw}{dv}=\frac{4+2w}{8-4w}=\frac{2+w}{4-2w}$$$$v\frac{dw}{dv}=\frac{2+w}{4-2w}-w=\frac{2-3w+2w^2}{4-2w}$$$$\frac{2-w}{2w^2-3w+2}dw=\frac{1}{2v}dv$$$$\int \frac{2-w}{2w^2-3w+2}dw=\int \frac{1}{2v}dv$$$$2\int \frac{1}{2w^2-3w+2}dw-\int \frac{w}{2w^2-3w+2}dw=\frac{1}{2}\ln v+C$$$$\sqrt{4\times 2\times 2-9}=\sqrt{7}$$$$\int \frac{1}{2w^2-3w+2}dw=\frac{2}{\sqrt{7}}{\tan }^{-1}\left(\frac{4w-3}{\sqrt{7}}\right)$$$$\int \frac{w}{2w^2-3w+2}dw=\frac{1}{4}\ln (2w^2-3w+2)+\frac{3}{2\sqrt{7}}{\tan }^{-1}\left(\frac{4w-3}{\sqrt{7}}\right)$$$$$$$$\frac{5}{2\sqrt{7}}{\tan }^{-1}\left(\frac{4w-3}{\sqrt{7}}\right)-\frac{1}{4}\ln (2w^2-3w+2)=\frac{1}{2}\ln v+C$$$$\frac{5}{\sqrt{7}}{\tan }^{-1}\left(\frac{4w-3}{\sqrt{7}}\right)-\frac{1}{2}\ln (2w^2-3w+2)=\ln v+C$$$$\frac{5}{\sqrt{7}}{\tan }^{-1}\left(\frac{4u-3v}{\sqrt{7}v}\right)-\frac{1}{2}\ln [2{\left(\frac{u}{v}\right)}^2-3\left(\frac{u}{v}\right)+2]=\ln v+C$$$$\frac{5}{\sqrt{7}}{\tan }^{-1}\left[\frac{4(y-\frac{11}{8})-3(x-\frac{1}{16})}{\sqrt{7}(x-\frac{1}{16})}\right]-\frac{1}{2}\ln [2{\left(\frac{y-\frac{11}{8}}{x-\frac{1}{16}}\right)}^2-3\left(\frac{y-\frac{11}{8}}{x-\frac{1}{16}}\right)+2]=\ln (x-\frac{1}{16})+C$$$$\frac{5}{\sqrt{7}}{\tan }^{-1}\left[\frac{4(16y-22)-3(16x-1)}{\sqrt{7}(16x-1)}\right]-\frac{1}{2}\ln [2{\left(\frac{16y-22}{16x-1}\right)}^2-3\left(\frac{16y-22}{16x-1}\right)+2]=\ln \left(\frac{16x-1}{16}\right)+C$$$$\frac{5}{\sqrt{7}}{\tan }^{-1}\left[\frac{64y-48x-85}{\sqrt{7}(16x-1)}\right]-\frac{1}{2}\ln [8{\left(\frac{8y-11}{16x-1}\right)}^2-6\left(\frac{8y-11}{16x-1}\right)+2]=\ln \left(\frac{16x-1}{16}\right)+C$$

Commented by tawa last updated on 17/Apr/17

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.$$

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