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(D) Given the differential equation: $(y^2+x+1) dy = (y+1) dx$. Rearranging the terms to form a linear differential equation in $x$: $\frac{dx}{dy} =

Vedclass · Accepted Answer

$\frac{x+2}{y+1}+\log (y+1)^2=y+c$

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(D) Given the differential equation: $(y^2+x+1) dy = (y+1) dx$. Rearranging the terms to form a linear differential equation in $x$: $\frac{dx}{dy} = \frac{y^2+x+1}{y+1} = \frac{y^2+1}{y+1} + \frac{x}{y+1}$. $\frac{dx}{dy} - \frac{1}{y+1} x = \frac{y^2+1}{y+1}$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y+1}$ and $Q(y) = \frac{y^2+1}{y+1}$. The integrating factor $IF = e^{\int P(y) dy} = e^{-\int \frac{1}{y+1} dy} = e^{-\log(y+1)} = \frac{1}{y+1}$. The solution is $x \cdot IF = \int Q(y) \cdot IF dy + c$. $x \cdot \frac{1}{y+1} = \int \frac{y^2+1}{(y+1)^2} dy + c$. Using the substitution $y^2+1 = (y+1)^2 - 2y - 1 + 1 = (y+1)^2 - 2(y+1) + 2$: $\frac{x}{y+1} = \int \left( 1 - \frac{2}{y+1} + \frac{2}{(y+1)^2} \right) dy + c$. $\frac{x}{y+1} = y - 2 \log |y+1| - \frac{2}{y+1} + c$. Multiplying by $(y+1)$: $x = y(y+1) - 2(y+1) \log |y+1| - 2 + c(y+1)$. Alternatively,rearranging the integral result: $\frac{x+2}{y+1} = y - 2 \log |y+1| + c$. Thus,$\frac{x+2}{y+1} + 2 \log |y+1| = y + c$,which is $\frac{x+2}{y+1} + \log (y+1)^2 = y + c$.

The general solution of the differential equation $(y^2+x+1) dy = (y+1) dx$ is

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