Question Number 91139 by john santu last updated on 28/Apr/20
$$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}+\mathrm{1}}{{y}−{x}+\mathrm{5}} \\ $$
Commented by john santu last updated on 28/Apr/20
![(dy/dx) = ((y−x+5−4)/(y−x+5)) = 1−(4/(y−x+5)) [ let Q=y−x ] [(dQ/dx) = (dy/dx)−1 ] ⇒(dQ/dx)+1 = 1−(4/(Q+5)) (dQ/dx) = −(4/(Q+5)) ⇒ (Q+5)dQ = −4dx ∫ (Q+5)dQ = −4x+C (1/2)(Q+5)^2 = C−4x Q+5 = ± (√(2C−8x)) y−x+5 = ± (√(2C−8x)) y = x−5±(√(2C−8x))](https://www.tinkutara.com/question/Q91145.png)
$$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}+\mathrm{5}−\mathrm{4}}{{y}−{x}+\mathrm{5}}\:=\:\mathrm{1}−\frac{\mathrm{4}}{{y}−{x}+\mathrm{5}} \\ $$$$\left[\:{let}\:{Q}={y}−{x}\:\right]\: \\ $$$$\left[\frac{{dQ}}{{dx}}\:=\:\frac{{dy}}{{dx}}−\mathrm{1}\:\right] \\ $$$$\Rightarrow\frac{{dQ}}{{dx}}+\mathrm{1}\:=\:\mathrm{1}−\frac{\mathrm{4}}{{Q}+\mathrm{5}}\: \\ $$$$\frac{{dQ}}{{dx}}\:=\:−\frac{\mathrm{4}}{{Q}+\mathrm{5}}\:\Rightarrow\:\left({Q}+\mathrm{5}\right){dQ}\:=\:−\mathrm{4}{dx} \\ $$$$\int\:\left({Q}+\mathrm{5}\right){dQ}\:=\:−\mathrm{4}{x}+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({Q}+\mathrm{5}\right)^{\mathrm{2}} =\:{C}−\mathrm{4}{x}\: \\ $$$${Q}+\mathrm{5}\:=\:\pm\:\sqrt{\mathrm{2}{C}−\mathrm{8}{x}}\: \\ $$$${y}−{x}+\mathrm{5}\:=\:\pm\:\sqrt{\mathrm{2}{C}−\mathrm{8}{x}} \\ $$$${y}\:=\:{x}−\mathrm{5}\pm\sqrt{\mathrm{2}{C}−\mathrm{8}{x}} \\ $$$$ \\ $$