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Question Number 168011 by Mastermind last updated on 31/Mar/22

$$Solve$$$$(2x+5y+1)dx-(5x+2y-1)dy=0$$$$$$$$Mastermind$$

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

$$\frac{dy}{dx}=\frac{2x+5y+1}{5x+2y-1}$$$$\frac{dy}{dx}=\frac{2(x+h)+5(y+k)}{5(x+h)+2(y+k)}$$$$2h+5k=1$$$$5h+2k=-1$$$$\Rightarrow h=-\frac{1}{3}$$$$\Rightarrow k=\frac{1}{3}$$$$letu=x+h=x-\frac{1}{3}$$$$v=y+k=y+\frac{1}{3}$$$$\frac{dv}{du}=\frac{2u+5v}{5u+2v}$$$$lett=\frac{v}{u}\Rightarrow v=ut$$$$\frac{dv}{du}=t+u\frac{dt}{du}$$$$t+u\frac{dt}{du}=\frac{2+5t}{5+2t}$$$$u\frac{dt}{du}=\frac{2(1-t^2)}{5+2t}$$$$\frac{(5+2t)dt}{1-t^2}=\frac{2du}{u}$$$$\frac{1}{2}\int (\frac{7}{1-t}+\frac{3}{1+t})dt=\int \frac{2du}{u}$$$$3\ln \mid 1+t\mid -7\ln \mid 1-t\mid =4\ln u+C$$$$3\ln \mid 1+\frac{y+\frac{1}{3}}{x-\frac{1}{3}}\mid -7\ln \mid 1-\frac{y+\frac{1}{3}}{x-\frac{1}{3}}\mid =4\ln u+C$$$$3\ln \mid 1+\frac{3y+1}{3x-1}\mid -7\ln \mid 1-\frac{3y+1}{3x-1}\mid =4\ln (x-\frac{1}{3})+C$$

Commented by Tawa11 last updated on 03/Apr/22

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

Commented by Mastermind last updated on 05/Apr/22

$$Thanksman$$

Answered by ajfour last updated on 06/Apr/22

$$\frac{2x+5y+1}{5x+2y-1}=\frac{dy}{dx}$$$$\frac{7x+7y}{3x-3y-2}=\frac{dy+dx}{dx-dy}$$$$x+y=s,x-y=t$$$$\frac{7s}{3t-2}=\frac{ds}{dt}$$$$\int \frac{ds}{s}=\frac{7}{3}\int \frac{3dt}{3t-2}$$$$\ln s=\frac{7}{3}\ln (3t-2)+c$$$$3\ln \mid x+y\mid =7\ln \mid 3x-3y-2\mid +3c$$$$$$

Commented by mr W last updated on 08/Apr/22

$$nicesolution!$$

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