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If gradient of a curve at any point P(x, y) is \frac{x+y+1}{2y+2x+1} and it is passes through origin, then curve is

Option: 1

2(x+3y)=ln\left | \frac{3x+3y+2 }{2}\right |


Option: 2

x+3y=ln\left | \frac{3x+3y+2 }{2}\right |


Option: 3

3y+x=ln\left (3x+2y+1 \right )


Option: 4

6y-3x=ln\left | \frac{3x+3y+2 }{2}\right |


Answers (1)

best_answer

 

Homogeneous Differential Equation -

Put

\frac{y}{x}=v

\frac{dy}{dx}=v+\frac{xdv}{dx}

-

 

 

Equations Reducible to the homogeneous form -

\frac{dy}{dx}=\frac{ax+by+c}{Ax+By+C}

- wherein

x=X+h

y=Y+k

 

 

 

\frac{dx}{dy}=\frac{(x+y)+1}{2(x+y)+1}

Put x+y=t, 1+\frac{dy}{dx}=\frac{dt}{dx}

\Rightarrow \frac{dt}{dx}=\frac{t+1}{2t+1}+1=\frac{3t+2}{2t+1}

\Rightarrow \int \frac{2t+1}{3t+2}dt=\int dx

\Rightarrow \frac{2t}{3}-\frac{1}{9}\ln(3t+2)=x+c

\Rightarrow 6(x+y)-\ln(3x+3y+2)=9x+c

\Rightarrow \ln(3x+3y+2)=6y-3x+c

Since it passes through (0,0) hence equation of curve is 6y-3x=ln\left | \frac{3x+3y+2}{2} \right |

Posted by

Deependra Verma

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