The general solution of the differential equa | vedclass

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(D) Given differential equation: $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$ Let $x+y=v$. Then $1+\frac{d y}{d x}=\frac{d v}{d x}$,so $\frac{d y}{d x}=\

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$\log _{e}|3 x+3 y+2|+3 x-6 y=C$

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(D) Given differential equation: $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$ Let $x+y=v$. Then $1+\frac{d y}{d x}=\frac{d v}{d x}$,so $\frac{d y}{d x}=\frac{d v}{d x}-1$. Substituting this into the equation: $\frac{d v}{d x}-1=\frac{v+1}{2 v+1}$ $\frac{d v}{d x}=\frac{v+1}{2 v+1}+1 = \frac{v+1+2 v+1}{2 v+1} = \frac{3 v+2}{2 v+1}$ Separating variables: $\frac{2 v+1}{3 v+2} d v=d x$ Rewrite the numerator: $\frac{\frac{2}{3}(3 v+2)-\frac{1}{3}}{3 v+2} d v=d x$ $\left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v=d x$ Integrating both sides: $\int \left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v = \int d x + C'$ $\frac{2}{3} v - \frac{1}{9} \log |3 v+2| = x + C'$ Substitute $v=x+y$: $\frac{2}{3}(x+y) - \frac{1}{9} \log |3 x+3 y+2| = x + C'$ Multiply by $9$: $6(x+y) - \log |3 x+3 y+2| = 9 x + 9 C'$ $6 x + 6 y - 9 x - \log |3 x+3 y+2| = C$ $-3 x + 6 y - \log |3 x+3 y+2| = C$ Multiplying by $-1$: $3 x - 6 y + \log |3 x+3 y+2| = C$ Thus,the correct option is $D$.

The general solution of the differential equation $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$ is

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